In article wG7Jb.96458$VB2.220705@attbi_s51, Loren Coe says...

i have been following you guys, haven't had the incentive to continue but
that will come, i have a spare bell xmfr and i know how to float a scope,
it has been awhile, tho. does anyone think the power meter would be of
any use (the one on the side of the home)?

I tried this as a first attempt to find the real power
for my idling phase converter motor. It didn't
work, there were enough other small loads in the house
that I was unwilling to shut off, that swamped
the 200 watt signal I was trying to see. If you could
be sure that *everything* else was off, you might
have a chance at it.

Jim

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Richard J Kinch wrote:

Back a ways you attempted to use a series regulator as an analogy.
This doesn't work since, in a series regulator, Iout ~= Iin and Vout
Vin.

No, Iout = Iin in a 3-terminal regulator, except for a small overhead
that runs the chip itself. Or, model it as a magic variable resistor if
you like.

What are you saying "No" to? I said Iout is approximately=equal=to Iin
and you said Iout=Iin except for ... Did you fail to understand "~="?
It's the closest approximation to the standard approximately-equal sign
available in ASCII and should be pretty obvious.

A switching regulator might be a better analogy: Vin*Iin ~=
Vout*Iout.

Hardly. A switching regulator is just a transformer, with an DC-AC
converter on the input, and AC-DC converter on the output. This would
be analogous to my imaginary "regulator" consisting of an air motor
regenerating the reservoir, not a conventional variable-restriction
regulator.

I see. Clearly you aren't to familiar with electronics terminology.
You are refering to a particular type of switching power supply in which
incomming DC is switched to provide AC which is then transformed and
rectified. Some use duty-cycle modulation to provide regulation, some
don't. The term switching regulator is usually applied to a device
where incoming DC is switched on and off into an L-C filter. The output
voltage is approximately the duty cycle times the input voltage. In
principle, neglecting switching and IR losses, the output power = input
power. In practice it can come pretty close. I've designed an built
switching regulators with over 95% efficiency.

Ted

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.

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.

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.

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.

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.

In article , Ted Edwards wrote:
jim rozen wrote:
But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

You could use another transformer - any handy 240 to whatever and put a
light load on it. Ground one side of the secondary and hook the other
side to the probe.

If it really is split 240 and not two lines of a three phase service,
looking at only one hot from ground should have very little error in
phase. Ted

this is a small 1960 tract home, most likely _not_ a two phase power
feed, the voltages for the low side is normally about 123vac, it varies
but have never seen it as low as 115 but as hi as 127. i suppose that
is not very definitive, the date s/b, here in Dallas-Plano. i always
thought the low side of a 208 feed was 110v?

we have had to install autoxmfrs in the early days for some equip that
needed 120vac. teletypes were especially picky. i gather the modern
3-ph feeds supply a higher voltage (as i read this ng). i remember some
real confusion in the late 60s, discussing two phase motors, many (my
father included) had strong opinions as to what they were (or were not).
we were just a bunch of computer repairmen, not EEs, and we got roped
in over our heads on some installations. the custormer and the (remote)
boss expected that we know more than they did.

it is easy to isolate the house load for short periods, so i will try
the power meter thing before dragging out the scope, probably tommorrow.
that s/b interesting, regardless. --Loren

In article , jim rozen wrote:
In article , Richard J Kinch
says...

But the compressor runs off of 240 volts. Don't you
need to put the scope across both hot leads?

No, you really just want to know the phase angle between the current versus
voltage. So any picture of the utility AC voltage will do (such as the
internal scope test signal!). You don't need to look directly at the
voltage supplied to the motor.

Not *just* any picture. The signal has to have exactly
the same phase as the incoming 240 volt line - because
when one is doing this, the power is mostly reactive,
so one is measuring a pretty small difference between
two large values. It would be worth checking to see
that the line phase reference really is good. Jim

okay, just how small? a few degrees? that would seem to obviate
the need for this experiment. Richard seems to be saying the internal
test signal is sync'd on the line frequency, more than a few microseconds
variation would be unexpected.

assuming this is a split phase supply (120/240), theoretically 180deg
phase shift would = 0 watts, right? i am expecting min of 30-60 degrees
on the scope which s/b plenty. and much more than that if indeed
the "compessed output" votes prove true.

of course i would check "calibration" to the internal signal before
measuring, just to be sure, but one less scope lead to my xfmr/resistor
kludge would be helpful. --Loren

In article , Ted Edwards wrote:
Also e-mailed due an ISP suffering Altsheimer's:
Loren Coe wrote:
does anyone have a suggestion as to how i measure this with
just "normal" equipment, including an O'scope? running back
and forth to the utility co.power meter doesn't seem like a
very good approach.

If your scope is dual trace, you can do what you want with a current
transformer. You need a 120V to 12V transformer rated for a dozen or so
amps. Connect a 10 Ohm, 10 Watt resistor across the 120V winding.
Connect the 12V winding in series with a motor lead. The "backwards"
transformer will have 11A flowing in the "12V" winding which will become
1.1A in the "120V" winding. 1.1A in a 10 ohm resistor is 11 volts so
the other winding (the "12V" winding) will drop only 1.1V. This will
have negligable effect on the motor operation. If you now pick up the
hot motor lead in one channel of the scope and the secondary of the
current transformer in the other channel, you can measure the time
difference between the waveforms and calculate the phase shift. Ted

can a higher resistance be used? this maybe a stupid question, but it
has been a _long_ time since basic electronics. does that reflect too
much impedance into the primary? duh, i think i know....

but riddle me this: won't i be more concerned with the delta than
the absolute phase relationship? the variation/180 = percentage
change in current thru the motor? if it is by some other factor,
then i need to know. 90deg? thanks! --Loren

jim rozen wrote:

have to use the correct voltage when figuring
I x V x cos(angle).

He said he had measured the current (11A) so presumably he has a clip on
ammeter. He also said, IIRC, he has a multimeter so can measure the
voltage across the motor. With the arrangement I suggested, he can
measure the time between zero crossings of the waveforms, dT. Then
angle=360*dT/16.7msec and we know V, I and angle. So
P=V x I x Cos(angle).

Ted

Loren Coe wrote:

this is a small 1960 tract home, most likely _not_ a two phase power
feed, the voltages for the low side is normally about 123vac, it varies
but have never seen it as low as 115 but as hi as 127. i suppose that
is not very definitive, the date s/b, here in Dallas-Plano. i always
thought the low side of a 208 feed was 110v?

To be sure, measure the hot to hot on the putative 240V as well as
neutral to each hot. If the sum of the hots to neutral equals (or is
very close to) the hot-to-hot, then it ain't two phases of a three phase
system.

Ted

jim rozen wrote:

In article wG7Jb.96458$VB2.220705@attbi_s51, Loren Coe says...

i have been following you guys, haven't had the incentive to continue but
that will come, i have a spare bell xmfr and i know how to float a scope,
it has been awhile, tho. does anyone think the power meter would be of
any use (the one on the side of the home)?

I tried this as a first attempt to find the real power
for my idling phase converter motor. It didn't
work, there were enough other small loads in the house
that I was unwilling to shut off, that swamped
the 200 watt signal I was trying to see. If you could
be sure that *everything* else was off, you might
have a chance at it.

Even if you can shut down _every_ load except the one in question, I
would want to ask just how accurate is the standard house meter at such
light loads. After all, they are concerend with an upper linmit on the
order of 20 or so KW. A few tens or hundreds of watts is a long way
below full scale.

Ted

In article , Ted Edwards says...

Even if you can shut down _every_ load except the one in question, I
would want to ask just how accurate is the standard house meter at such
light loads. After all, they are concerend with an upper linmit on the
order of 20 or so KW. A few tens or hundreds of watts is a long way
below full scale.

Agree. Really the best way is to use a floating scope
like that scope meter, and a sampling resistor, to
display both current and voltage, and then measure
the phase angle.

Jim

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In article , Ted Edwards says...

measure the time between zero crossings of the waveforms, dT. Then
angle=360*dT/16.7msec and we know V, I and angle. So
P=V x I x Cos(angle).

Right, my point was that if he's measuring the phase
angle of a 120 volt waveform, but the compressor
is running on 240, he needs to plug the 240 number
into your calculation above. Not the 120.

Jim

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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

In article ,
says...
If I might take a moment to dither about the semantics
here.

"lowering the potential to do mechanical work" is
a bit different in some sense, than "loss." Because
of the history of electrical analogy here, this may
be confusing.

Yeah, "loss" is not strictly appropriate, but I don't think
it has caused too much confusion in the context of this
thread, and there's been worse use of terminology. In
particular, "potential energy" has been much misused.


Most folks think of the term 'loss' as a point loss,
ie a lumped circuit element that is dissipative. The
regulator, or the connection between the two tanks
in the case of no regulator, or even the expansion
orifice in the case where gas is expanding into
vacuum, are not analogous to electrical lumped circuit
elements like resistors. No heat appears in them
during the process. [1]

I've heard the term "lumped", but don't have even a vague
notion of what it means. I assume it's a simple concept?


(electrical analog discussion really are
doomed to fail here...)

Sure the end result having two tanks with three times
the volume, filled at a lower pressure, has less stored
potential. But the difference does not appear in the
regulator. The energy was not lost in the regulator.

A hydraulic regulator is functionally the same as a
pneumatic regulator, but it does get hot in operation.
Would you say that energy was lost in the regulator in that
case? (Again with the sloppy language.)

Ned Simmons

In article , Ned Simmons
says...

I've heard the term "lumped", but don't have even a vague
notion of what it means. I assume it's a simple concept?

Yes, the idea is 'lumped' vs 'distributed' circuit
elements. For example, you could imagine a circuit
that has a battery, a wire that leads to a resistor,
and then another wire that connects back to the battery.

In most cases, almost all the power shows up in one
spot - dissipated in the resistor.

Now change the setup a bit to have the resistor hooked
up to the battery with nichrome wire. Now some power
shows up in the resistor, and some shows up in the
wires themselves. It makes sense to look at that second
type as more of a distributed element load.

You can also see the same effect with a tuned cicuit
that consists of a single capacitor and coil, hooked
up with traces on a circuit board. If the tuning
frequency is low enough, the circuit board traces
are simply 'wires' and don't have to be considered.
This is a lumped element circuit.

Then make the tuning frequency very high, and the
circuit board traces become part of the tuning
circuit - with their capacitance and inductance
contributing to the overall tuning. At some point,
the traces themselves become the tuned circuit
elements. So at high frequencies it's a distributed
circuit.

My point was, looking at only one location of the
regulator/tanks setup will lead to confusing
conclusions. The overall system will obey conservation
of energy laws if all the energy can be accounted for.
This includes things like the internal thermal energy
of the gas, any mechanical work that is done, the
kinetic energy of the flowing gas, and whatever potential
energy is still stored in the compressed gas in various
locations.

A hydraulic regulator is functionally the same as a
pneumatic regulator, but it does get hot in operation.
Would you say that energy was lost in the regulator in that
case? (Again with the sloppy language.)

Hydraulic fluid of course is incompressible. But as
long as the regulator isn't getting hot simply because
the fluid flowing through it is hot, then I would
tend to say that yes, that would be a sign that
there is a sort of lumped element loss at that point
in the hydraulic circuit.

Jim

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On Fri, 02 Jan 2004 01:21:40 -0600, Richard J Kinch
wrote:

Don Foreman writes:

Something can get colder and gain heat at the same
time.

There's a law of thermodynamics that says otherwise.

Compressed air exiting an orifice gets colder (temperature decreases) from
expansion, while gaining some new heat from the friction and turbulence of
passing through the orifice. Which law of thermodynamics "says otherwise"?

Perhaps the one that sez Foreman maybe oughtta stick to electronics?

I've always thought of temperature as proportional to heat content in
joules or calories times specific heat of the material holding it.

In article , Ned Simmons
says...

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.

I think *that* question is pretty well answered.

The point is seen in reverse, when one realizes that
any attempt to illustrate electrical principles using
the old 'pump, tank, pipe, valve' teaching tools
does indeed have certain distinct, um, limitations.

Jim

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In article , Don Foreman says...

Something can get colder and gain heat at the same
time.

I've always thought of temperature as proportional to heat content in
joules or calories times specific heat of the material holding it.

For this discussion about ideal gasses, the temperature of the
gas is proportional to the internal energy of the atoms, by
a simple proportionality.

E = 3/2 x K x T

The K is the boltzman constant (guy who figured this out),
T is the temperature in degrees kelvin (so-called absolute)
and the 3/2 is there because each degree of mechanical
freedom of an individual gas atom has 1/2 kT of energy.
Because it can move in X, Y, and Z, the total is 3/2.

I think what you were saying was, that there could be
*two* things going on in the regulator at one time.
The first being expansion of the gas, which causes
it to cool. The second being the friction or turbulence
inside the regulator, which causes the gas to heat.

The *net* effect might still be an overall cooling
of the regulator, even though there is some heating
going on from the friction losses. The cooling is
just larger than the heating.

Jim

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On Sat, 3 Jan 2004 00:20:35 -0500, Ned Simmons
wrote:

I just did the same in Mathcad and am getting the same
results. (But shouldn't your units be inch-pounds?)

Oops! Yes, inch-pounds.

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

OK. I'll try to document it so it's clear what folly I've wrought in
the form of equations. :)


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.

That was the question. I think it's clear that they are not
analogous. One must keep track of all of the energy, not just
pressure and flowrate. That was the flaw in my treating power as
pressure * flowrate. Power must be treated as rate of flow of total
energy.

The fact that a regulator doesn't get noticably warm indicates to me
that power (and energy) is not being lost there, but I'm still not
clear on how that can be. My best guess at the moment is that some
potential energy is converted from pressure to heat but that the heat
is swept away in the downstream gas so the energy is preserved in the
downstream gas.

It might be interesting to put a thermocouple on a regulator, wrap the
lot in thermal insulation, run it for a while and see what happens.

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

Roger that!

Ned Simmons

In article ,
says...
In article , Ned Simmons
says...

I've heard the term "lumped", but don't have even a vague
notion of what it means. I assume it's a simple concept?

Yes, the idea is 'lumped' vs 'distributed' circuit
elements. For example, you could imagine a circuit
that has a battery, a wire that leads to a resistor,
and then another wire that connects back to the battery.

In most cases, almost all the power shows up in one
spot - dissipated in the resistor.

Now change the setup a bit to have the resistor hooked
up to the battery with nichrome wire. Now some power
shows up in the resistor, and some shows up in the
wires themselves. It makes sense to look at that second
type as more of a distributed element load.

You can also see the same effect with a tuned cicuit
that consists of a single capacitor and coil, hooked
up with traces on a circuit board. If the tuning
frequency is low enough, the circuit board traces
are simply 'wires' and don't have to be considered.
This is a lumped element circuit.

Then make the tuning frequency very high, and the
circuit board traces become part of the tuning
circuit - with their capacitance and inductance
contributing to the overall tuning. At some point,
the traces themselves become the tuned circuit
elements. So at high frequencies it's a distributed
circuit.

My point was, looking at only one location of the
regulator/tanks setup will lead to confusing
conclusions. The overall system will obey conservation
of energy laws if all the energy can be accounted for.
This includes things like the internal thermal energy
of the gas, any mechanical work that is done, the
kinetic energy of the flowing gas, and whatever potential
energy is still stored in the compressed gas in various
locations.

OK, I see what you're saying. I think it's a matter of
perspective and the level of understanding you're shooting
for.

For the example I was using - the isothermal
reservoir/regulator/piston system - saying, "the loss
occurs in the regulator" doesn't really have any bearing on
the solution of the problem. However, if you're trying to
understand why the regulator gets colder, saying "the loss
occurs in the regulator" obscures the explanation. In one
case you're looking at the process as a whole, in the other
you're trying to break that process down into smaller
chunks. Is that your point?

Similarly, in the electical analogy, wouldn't you
eventually want to abandon the concept of power loss in a
resistor when looking for a very fundamental understanding
of the loss?


A hydraulic regulator is functionally the same as a
pneumatic regulator, but it does get hot in operation.
Would you say that energy was lost in the regulator in that
case? (Again with the sloppy language.)

Hydraulic fluid of course is incompressible. But as
long as the regulator isn't getting hot simply because
the fluid flowing through it is hot, then I would
tend to say that yes, that would be a sign that
there is a sort of lumped element loss at that point
in the hydraulic circuit.


The oil gets hot when it loses pressure in the regulator,
so even though the oil entering the regulator may be at
ambient, the oil heats the regulator. Of course, unlike
electric current flowing thru a resistor, the oil also
carries heat away. Another chink in the electrical analogy.

Ned Simmons

In article ,
says...
In article , Ned Simmons
says...

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.

I think *that* question is pretty well answered.

Care to wager when it'll come up again. g


The point is seen in reverse, when one realizes that
any attempt to illustrate electrical principles using
the old 'pump, tank, pipe, valve' teaching tools
does indeed have certain distinct, um, limitations.


Yup. As I said in my first post in this thread, "any
analogy between electricity and a compressible gas is
doomed." Comparisons to noncompressible flow can get you a
little farther, but have to be abandoned at some point as
well.

Ned Simmons

In article , Ned Simmons
says...


Yup. As I said in my first post in this thread, "any
analogy between electricity and a compressible gas is
doomed."

And yet you see them, ALL the time. I think this
is because electricity used to be taught that way,
with the pipes/valves/tanks/pumps. Some folks
simply cannot wrap their minds around electricity
so the plumbing analogies show up in the older texbooks.

My dad a case in point. He understands how electricity
flows. But he never did grasp alternating current.
How can it do any work if it's only going back and
forth?

Jim

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On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons wrote:
In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch wrote:


In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the base
of a dam, the other 999,999,999,999,999 gallons in the reservoir are
wasted. Obviously not true at all. You use what you use, the rest stays
in the reservoir (with all of its potential energy intact) until you need it.


Another faulty analogy.

Not an analogy. The man said regulation of *any power source* and
specifically listed "water behind a dam". Not all regulators are shunt
regulators which waste any energy their downstream load does not
demand. Most regulation is obtained simply by not allowing any more
energy than the downstream load requires to be released from the
reservoir, be that water behind a dam, the energy stored in a flywheel,
or the energy stored in a capacitor, or a compressed air tank.


Since water is essentially
incompressible, the capacity of the water behind a dam to
do mechanical work is almost entirely due to its potential
energy -- its elevation in a gravitational field.

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

It is easy to illustrate that if you're just venting air to atmosphere,
but it is wrong if you're requiring the air to do work beyond the
regulator. The latter is the normal case except when you're just
bleeding down the tank.

Example, assume an air cylinder with a piston surface area of
12 square inches and a stroke of 1 foot. If you feed that cylinder
50 PSI air regulated from a 60 gallon 175 PSI tank, you can do
600 ft-lbs of work by lifting a 600 pound weight the 1 foot stroke
of the cylinder. Now vent the cylinder and fill it again from the
regulated source. You do another 600 ft-lbs of work. Do it again,
and again. You can keep doing it until the pressure in the tank
falls below 50 PSI.

Now try it without the regulator. You can't do it as many times,
because energy stored in the tank has been *wasted* at each fill,
ie the pressure in the cylinder and tank equalize each time you
open the fill valve even though that's more energy than you *need*
to lift the weight one foot, and that excess energy is lost when you
vent the cylinder. This is a clear case where the *absence* of a
regulator wastes energy. It is in fact the *usual* case.

Gary

On Wed, 31 Dec 2003 23:45:21 -0600, Richard J Kinch wrote:
Don Foreman writes
The loss
is at the compressor, not at the regulator. The compressor gets hot
but the regulator does not, right?

The regulator can get downright chilly. Temperature is not heat or a
measure of heat. Something can get colder and gain heat at the same
time.

It cannot if that something has a fixed mass of material with a fixed heat
capacity. I think it is safe to say that any given regulator has a fixed mass,
and the material of which it is made does not change during operation, so
the heat capacity does not change. Therefore, the heat in the regulator
*is* directly proportional to the regulator temperature.

Gary

On Mon, 29 Dec 2003 01:08:00 -0600, Richard J Kinch wrote:
Gary Coffman writes:

Power doesn't go anywhere. It is a *rate*, not an actual thing that
moves.

You make a contemptibly false statement, ignorant of even high school
physics.

Perhaps some high school physics classes leave their students ignorant
enough to believe your claim. But my university degree in physics says
you're wrong. A *rate* is a measure of how quickly *something* else is
converted, moved, or otherwise transferred past an observation point.
It is that *something else*, in this case energy, which does the moving.
Power is merely an observation of that energy flow past a fixed point.

Consider this, if power can flow, what are the units of that flow?
It would have to be a rate of a rate, ie kg-m/sec/sec, and that
isn't the units of a watt. What sort of meter would you need to
measure it?

Gary

In article , Gary Coffman says...

Power is merely an observation of that energy flow past a fixed point.

I guess I have a problem with the 'fixed point' definition.
This sounds too much like the Poynting Vector.

If there is a battery, and a resistor hooked up with wires
to the battery, the resistor gets hot.

The thing going by through the wires, past a fixed point,
are so-many coulombs per second of electrons. The
power dissipates in the resistor, but even that is
an extended object. More so if one uses nichrome
wire rather than regular wire, and no other resistor
at all.

Work done, or energy converted, per unit time might
be a good general approach.

Jim

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In article , Gary Coffman wrote:
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons wrote:
In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch wrote:
In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.
.....
50 PSI air regulated from a 60 gallon 175 PSI tank, you can do
600 ft-lbs of work by lifting a 600 pound weight the 1 foot stroke
of the cylinder. Now vent the cylinder and fill it again from the
regulated source. You do another 600 ft-lbs of work. Do it again,
and again. You can keep doing it until the pressure in the tank
falls below 50 PSI.

Now try it without the regulator. You can't do it as many times,
because energy stored in the tank has been *wasted* at each fill,
ie the pressure in the cylinder and tank equalize each time you
open the fill valve even though that's more energy than you *need*
to lift the weight one foot, and that excess energy is lost when you
vent the cylinder. This is a clear case where the *absence* of a
regulator wastes energy. It is in fact the *usual* case. Gary

i don't propose to add anything, just observe that US High School
Phyics covered this well. the difference in hp vs work vs energy.
the whole concept of air regulator losses seems to be off topic,
but then i cannot retrieve the orig. post.

there is nothing in a gas regulator that _can_ dissapate significant
enery, imho. and in _normal_ operation, it is one of the cheapest, lowest
overhead controls invented by man. if you want to assign the loss of energy
to the orifice that vents it, we come from different schools. --Loren

Gary Coffman wrote:
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons
wrote:
In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch
wrote:


In general, regulation of *any power source* implies an inherent
waste versus the unregulated source: electrical power, compressed
air power, motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the
base of a dam, the other 999,999,999,999,999 gallons in the
reservoir are wasted. Obviously not true at all. You use what you
use, the rest stays in the reservoir (with all of its potential
energy intact) until you need it.


Another faulty analogy.

Not an analogy. The man said regulation of *any power source* and
specifically listed "water behind a dam". Not all regulators are shunt
regulators which waste any energy their downstream load does not
demand. Most regulation is obtained simply by not allowing any more
energy than the downstream load requires to be released from the
reservoir, be that water behind a dam, the energy stored in a
flywheel, or the energy stored in a capacitor, or a compressed air
tank.


Since water is essentially
incompressible, the capacity of the water behind a dam to
do mechanical work is almost entirely due to its potential
energy -- its elevation in a gravitational field.

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

It is easy to illustrate that if you're just venting air to
atmosphere, but it is wrong if you're requiring the air to do work
beyond the regulator. The latter is the normal case except when
you're just bleeding down the tank.

The final regulated pressure has to be matched to the task, I don't think
anyone would disagree with that. However, the pressure supplied to the
regulator should allow for some headroom and optimally no more.



Example, assume an air cylinder with a piston surface area of
12 square inches and a stroke of 1 foot. If you feed that cylinder
50 PSI air regulated from a 60 gallon 175 PSI tank, you can do
600 ft-lbs of work by lifting a 600 pound weight the 1 foot stroke
of the cylinder. Now vent the cylinder and fill it again from the
regulated source. You do another 600 ft-lbs of work. Do it again,
and again. You can keep doing it until the pressure in the tank
falls below 50 PSI.

Now try it without the regulator. You can't do it as many times,
because energy stored in the tank has been *wasted* at each fill,
ie the pressure in the cylinder and tank equalize each time you
open the fill valve even though that's more energy than you *need*
to lift the weight one foot, and that excess energy is lost when you
vent the cylinder. This is a clear case where the *absence* of a
regulator wastes energy. It is in fact the *usual* case.

Gary

I don't think anybody is arguing against the use of regulators. The point
that a few of us have been trying to make is that x liters of air at y
pressure can do more work than 2x liters of air at y/2 pressure. Double the
pressure and you double both the force and the distance that the force can
be applied, in the case of the piston that Ned cited. The original post
surmised that a regulator was analogous to a transformer. It's not. A gas
regulator also cannot be compared to regulating water from a tank.

On 4 Jan 2004 17:24:06 -0800, jim rozen wrote:
In article , Gary Coffman says...

Power is merely an observation of that energy flow past a fixed point.

I guess I have a problem with the 'fixed point' definition.
This sounds too much like the Poynting Vector.

If there is a battery, and a resistor hooked up with wires
to the battery, the resistor gets hot.

The thing going by through the wires, past a fixed point,
are so-many coulombs per second of electrons. The
power dissipates in the resistor, but even that is
an extended object. More so if one uses nichrome
wire rather than regular wire, and no other resistor
at all.

Power doesn't dissipate. Energy is converted from
electrical to thermal in the resistor. Power is the rate
of the conversion.

Gary

In article , Gary Coffman says...

Power doesn't dissipate.

Right you are, of course.
Another definition mine field!

Jim

==================================================
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In article , jim rozen wrote:
In article , Ned Simmons
says...
Yup. As I said in my first post in this thread, "any
analogy between electricity and a compressible gas is
doomed."

And yet you see them, ALL the time. I think this
is because electricity used to be taught that way,
with the pipes/valves/tanks/pumps. Some folks
simply cannot wrap their minds around electricity
so the plumbing analogies show up in the older texbooks.

My dad a case in point. He understands how electricity
flows. But he never did grasp alternating current.
How can it do any work if it's only going back and
forth? Jim

i am kinda like your Dad, i used to think i knew about
basic electronics and current flow, ac or dc. but when
they decided that "holes" were the current and reversed
the flow i kinda lost orientation.

the old electron flow theory has served me well, however,
at the level i have applied my knowledge. --Loren

In article ,
says...
On Sun, 28 Dec 2003 23:56:33 -0500, Ned Simmons wrote:
In article ,
says...
On Sat, 27 Dec 2003 00:50:14 -0600, Richard J Kinch wrote:


In general, regulation of *any power source* implies an inherent waste
versus the unregulated source: electrical power, compressed air power,
motive power, water behind a dam, etc.

So, if you draw 1 gallon of water a minute through a valve at the base
of a dam, the other 999,999,999,999,999 gallons in the reservoir are
wasted. Obviously not true at all. You use what you use, the rest stays
in the reservoir (with all of its potential energy intact) until you need it.


Another faulty analogy.

Not an analogy. The man said regulation of *any power source* and
specifically listed "water behind a dam". Not all regulators are shunt
regulators which waste any energy their downstream load does not
demand. Most regulation is obtained simply by not allowing any more
energy than the downstream load requires to be released from the
reservoir, be that water behind a dam, the energy stored in a flywheel,
or the energy stored in a capacitor, or a compressed air tank.

OK, let's look at the water behind a dam. If you remove a
gallon of water from the reservoir you lower the potential
energy of the remaining mass of water. The reduction in PE
is the same whether you:

A. remove the water under pressure at the bottom of the dam
and generate electricity,

B. remove the water under pressure at the bottom of the
dam, regulate the pressure down and generate less
electricity,

C. bypass the generator completely and spray the water into
the air, or

D. dip the water out of the top of the reservoir in a
bucket and spill it over the other side of the dam.

In each case the decrease in the mass of the water and the
change in water level (elevation in the gravitational
field) is the same, thus the delta PE is the same in all
four cases.



Since water is essentially
incompressible, the capacity of the water behind a dam to
do mechanical work is almost entirely due to its potential
energy -- its elevation in a gravitational field.

It's easy to demonstrate, without resorting to
thermodynamics, that air expanding thru a regulator loses
some of its capacity to do mechanical work, and that energy
does *not* remain in the air behind the regulator.

It is easy to illustrate that if you're just venting air to atmosphere,
but it is wrong if you're requiring the air to do work beyond the
regulator. The latter is the normal case except when you're just
bleeding down the tank.

If that were true you'd be able to determine how much
mechanical work had been done by the air that has left the
tank by simply monitoring the gas in the tank. Do you
really think any data you gather *in the tank* will tell
whether the air that has *left the tank* was used to
perform mechanical work, or simply released to the
atmosphere?

The same aplies in the case of the water in the reservoir
above.


Example, assume an air cylinder with a piston surface area of
12 square inches and a stroke of 1 foot. If you feed that cylinder
50 PSI air regulated from a 60 gallon 175 PSI tank, you can do
600 ft-lbs of work by lifting a 600 pound weight the 1 foot stroke
of the cylinder. Now vent the cylinder and fill it again from the
regulated source. You do another 600 ft-lbs of work. Do it again,
and again. You can keep doing it until the pressure in the tank
falls below 50 PSI.

Now try it without the regulator. You can't do it as many times,
because energy stored in the tank has been *wasted* at each fill,
ie the pressure in the cylinder and tank equalize each time you
open the fill valve even though that's more energy than you *need*
to lift the weight one foot, and that excess energy is lost when you
vent the cylinder. This is a clear case where the *absence* of a
regulator wastes energy. It is in fact the *usual* case.

This is like saying there is no loss in a shunt regulator
because it can be used to supply the proper voltage to a
device that may self destruct or draw excessive current in
the absence of the regulator. (Just had to work in an
electrical analogy.)

Ned Simmons

Ted Edwards writes:

What are you saying "No" to? I said Iout is approximately=equal=to
Iin and you said Iout=Iin except for ... Did you fail to understand
"~="? It's the closest approximation to the standard
approximately-equal sign available in ASCII and should be pretty
obvious.

Yes, I misunderstood. The typeset math symbol you refer to does not
contain a tilde, nor does a tilde mean anything in mathematics. The
ASCII tilde as an operator in the C programming language explicitly
means "not" or "negation", and thus I understood tilde-equals to mean
not-equals.

A switching regulator might be a better analogy: Vin*Iin ~=
Vout*Iout.

Hardly. A switching regulator is just a transformer, with an DC-AC
converter on the input, and AC-DC converter on the output. This
would be analogous to my imaginary "regulator" consisting of an air
motor regenerating the reservoir, not a conventional
variable-restriction regulator.

I see. Clearly you aren't to familiar with electronics terminology.

Clearly: http://www.truetex.com/resume.pdf

It doesn't even matter what you call your electronic device. The
switching regulator
analogy just doesn't fit. Chopping current (a charge pump) is not like
expanding air (via a restriction).

I've designed an built switching regulators with over 95% efficiency.

Which proves they are not analogous to a variable-restriction air-
pressure regulator, something inherent inefficient.

jim rozen writes:

I think what you were saying was, that there could be
*two* things going on in the regulator at one time.
The first being expansion of the gas, which causes
it to cool. The second being the friction or turbulence
inside the regulator, which causes the gas to heat.

Yes.

In the extreme, I imagine if you ran an air motor into a brake, powered by
a can of boiling liquid freon, you would wind up with a lower temperature,
yet more heat.

On Mon, 5 Jan 2004 20:25:10 -0500, Ned Simmons wrote:
In article ,
says...
Example, assume an air cylinder with a piston surface area of
12 square inches and a stroke of 1 foot. If you feed that cylinder
50 PSI air regulated from a 60 gallon 175 PSI tank, you can do
600 ft-lbs of work by lifting a 600 pound weight the 1 foot stroke
of the cylinder. Now vent the cylinder and fill it again from the
regulated source. You do another 600 ft-lbs of work. Do it again,
and again. You can keep doing it until the pressure in the tank
falls below 50 PSI.

Now try it without the regulator. You can't do it as many times,
because energy stored in the tank has been *wasted* at each fill,
ie the pressure in the cylinder and tank equalize each time you
open the fill valve even though that's more energy than you *need*
to lift the weight one foot, and that excess energy is lost when you
vent the cylinder. This is a clear case where the *absence* of a
regulator wastes energy. It is in fact the *usual* case.

This is like saying there is no loss in a shunt regulator
because it can be used to supply the proper voltage to a
device that may self destruct or draw excessive current in
the absence of the regulator. (Just had to work in an
electrical analogy.)

No! It is not even remotely similar. A shunt regulator accepts
*every bit of energy the source is capable of supplying* at any
given moment, passes part of it to the load, and converts the
rest of it to heat in the shunt resistance.

An air regulator valves *just enough* energy from the tank to
satisfy the load at any given moment. No more. The rest
*remains in the tank*.

A *very small* amount of energy may be converted to heat
in the regulator valve due to friction and gas turbulence,
but for the most part the regulator valve acts like the
dissipationless dynamic resistance of a vacuum tube,
limiting the amount of energy passing through it under
feedback control, but not dissipating any excess the power
supply might have been able to supply. The latter energy
remains behind in the power supply capacitors/transformer-
rectifier/mains/ and ultimately in the generating station fuel
supply.

Gary

Don Foreman writes:

I've always thought of temperature as proportional to heat content in
joules or calories times specific heat of the material holding it.

If that were always true, you couldn't have air conditioning, or an ice
cold drink.

Stretch a fat rubber band. What happens to the temperature of the middle
of it (use your perioral sensor)? Let it equilibrate to the ambient air
for a moment, and then relax it, and then what happens to the temperature?
Temperature and heat content are not always related.

Loren Coe writes:

Richard seems to be saying the internal
test signal is sync'd on the line frequency, more than a few microseconds
variation would be unexpected.

Correct, this is the point of having it as a trigger reference, to see what
devices are doing relative to the wall outlet.

assuming this is a split phase supply (120/240), theoretically 180deg
phase shift would = 0 watts, right?

I would expect it to be the absolute value of the cosine, which is zero at
90 and 270, and 100 percent at 0 and 180.

Ned Simmons writes:

Of course, unlike
electric current flowing thru a resistor, the oil also
carries heat away. Another chink in the electrical analogy.

Every analogy has limits. The point of this one was simply that losses
occur and result in waste heat, and the current (volume) and voltage
(pressure) behave accordingly. And you can buy junky ones at Radio Shack
(Harbor Freight).

jim rozen writes:

Sure the end result having two tanks with three times
the volume, filled at a lower pressure, has less stored
potential. But the difference does not appear in the
regulator. The energy was not lost in the regulator.

I'll agree to this as your definitional retreat.

Ted Edwards writes:

After all, they are concerend with an upper linmit on the
order of 20 or so KW. A few tens or hundreds of watts is a long way
below full scale.

I always thought they could track a night light. They run at incredibly
slow rpms.

Don Foreman writes:

Could a regulator delivering a couple of HP worth of air be
dissipating a significant fraction of an HP as heat?

It has to. You can regulate any flow down to arbitrarily close to 0 psig,
like a big sintered diffuser. All the energy is spent.

Remember, besides delivering work to the tool or heat to the regulator (or
regulator output if you like), expanding air also does work against the
atmosphere. When you compress free air, you shrink the Earth's atmosphere
(!) by that much, and blow it up again when you expand.