Don Foreman writes:

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.

Did you ever pop open the overpressure relief valve, and let it run for a
few minutes to bleed a full tank? No thermocouple needed to feel that
result.

Ted Edwards writes:

P=V x I x Cos(angle)

P=V x I x abs(Cos(angle)) ?

Gary Coffman writes:

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.

You are correct that with-the-regulator beats without-the-regulator, but
compressing-the-air-to-50-psi-to-start-with beats them both, which is to
say, if mechanical efficiency is the figure of merit, then a regulator on a
high-pressure supply is wasteful compared to an unregulated source at the
application pressure.

Gary Coffman writes:

Therefore, the heat in the regulator
*is* directly proportional to the regulator temperature.

The system in view consists of the regulator and the air, not just the
regulator. If the air expands, this system can decrease in temperature
while gaining heat. And furthermore, the air flow can segregate the heat
and temperatures, so some parts of the system get colder and some hotter
(in temperature).

Gary Coffman writes:

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.

You say that a "rate" is "not an actual thing that moves". By your logic,
then "energy" is not a thing either, it is just a rate at which molecules
pass by at some density.

When you get down to the philosophical underpinnings, a "thing" in physics
is just whatever you can measure, which is to say, something sensible in
any quantitative manner, whether by direct human sense or via intermediate
instruments.

Do you believe in electrons, or in holes, or both, or what?

Gary Coffman writes:

Power doesn't dissipate.

Hitler had power, and was dissipated. I hereby invoke Godwin's Law to
close this thread.

Gary Coffman writes:

dissipationless dynamic resistance

Get a patent on that.

In article , Richard J Kinch
says...

I'll agree to this as your definitional retreat.

"Defintional Retreat?"

That's where the terms and phrases used have a
precise meaning, in accordance with accepted
principles of science.

I can live with that.

Jim

==================================================
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==================================================

In article , Richard J Kinch
says...

Temperature and heat content are not always related.

In an ideal gas, they are. Another definitional
retreat - or simply the contents of an introductory
textbook? I'll let you make the call on that.

Jim

==================================================
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==================================================

In article ,
says...

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.


All right. Yet another bad electrical analogy. I oughta
know better by now.

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*.

Then please respond to my previous post and explain how
you're going to determine whether the water removed from a
reservoir, or the air released from a pressurized tank, was
used to do mechanical work, or simply released to the
environment. If you're correct it will be possible to tell
without looking beyond the boundaries of the reservoir or
tank.

Ned Simmons

"Gary Coffman" wrote in message
...
dissipationless dynamic resistance of a vacuum tube,

So why do my plates glow when they are overbiased? Do you recommend
I grab onto a 6L6 for more than 30 seconds?

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

On Tue, 06 Jan 2004 03:24:22 -0600, Richard J Kinch
wrote:

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.

You're right. I was thinking only of the metal in the regulator when
we said "the regulator gets cold".

However, if the regulator is colder than ambient then it is not
dissipating energy as heat. (If anything it's recovering some heat
from the environment, some of which was put there at the compressor
during compression.) If the regulator isn't losing energy to the
environment in the form of heat, then all energy coming in with the
high pressure gas must be going out with the lower-pressure gas --
so where's the loss in the regulator?

On Tue, 06 Jan 2004 04:04:40 -0600, Richard J Kinch
wrote:

Gary Coffman writes:

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.

You are correct that with-the-regulator beats without-the-regulator, but
compressing-the-air-to-50-psi-to-start-with beats them both, which is to
say, if mechanical efficiency is the figure of merit, then a regulator on a
high-pressure supply is wasteful compared to an unregulated source at the
application pressure.

Yes, at the system level. It still has NOT been shown how any
energy remains in the regulator, or leaves it other than in the
downstream airflow, if the regulator doesn't get warmer. If energy
out = energy in, then there's no net loss at the regulator. We've
looked at other parts of the system and thought we accounted for the
energy and so inferred or deduced that there must be loss in the
regulator, but we've all failed to show how any energy leaves the
regulator other than in the gas going to the point of use.

Geez, my regulator is only 2 feet from me. Hope it ain't emitting
Xrays!

Don Foreman writes:

If energy
out = energy in, then there's no net loss at the regulator.

Sez who? The "loss" is in the conversion of some of the useful-energy-in
into waste-heat-out and entropy-out. Whether this waste heat stays in
the regulator to raise its temperature, versus being carried downstream,
doesn't change the fact that the regulator directly caused the waste.

Remember that part of the loss is also in the form of work performed on the
ambient atmosphere, not heat.

On Thu, 08 Jan 2004 01:13:24 -0600, Richard J Kinch
wrote:

Don Foreman writes:

If energy
out = energy in, then there's no net loss at the regulator.

Sez who? The "loss" is in the conversion of some of the useful-energy-in
into waste-heat-out and entropy-out. Whether this waste heat stays in
the regulator to raise its temperature, versus being carried downstream,
doesn't change the fact that the regulator directly caused the waste.

Remember that part of the loss is also in the form of work performed on the
ambient atmosphere, not heat.

What waste? Heat is still energy. If the downstream air is colder
than ambient then heat is not lost by conduction or radiation to
ambient, so that energy is still in the downstream gas.

Maybe we're getting closer to understanding, though. We've agreed
that a certain amount (mass) of air is compressed, and some heat is
lost in that process. Compression to higher pressure results in more
work being converted to heat (usually lost) at the compressor or in
the reservoir as that air cools by conduction to ambient -- and
perhaps a little radiation as well. But what about after that, as
in from just before the regulator to point of exhaust? I think that
was the original question.

Adding heat from the regulator to the air downstream of the
compressor will raise it's temperature above what it would otherwise
be, hence it's volume at regulated pressure, so less high-pressure air
will be required per unit volume of air at regulated pressure.

However, air exhausting to atmosphere from the point of use will
carry heat with it, the amount of heat depending on it's temperature.
The warmer the exhaust gas is (even if below ambient temp), the more
heat (hence energy) is lost by that route. Maybe *that's* where the
thus-far unaccounted-for energy goes.

I'll admit to being a bit disappointed that you, with your impressive
academic credentials, didn't clear this up many messages ago. It was
said somewhere along the line that I am incredibly stubborn. That's
true. I don't care a bit about "winning the argument" or "being
the one that's right" but I'm admittedly stubborn as hell about
persisting until I understand and continuing to question "expert
exhortations" until I understand. I have the greatest respect
for those who will teach and share, am not a bit impressed with
wizards who may know but can't or won't help others to know and
understand as well.

Perhaps my age is showing?

On Tue, 06 Jan 2004 04:31:22 -0600, Richard J Kinch wrote:
Gary Coffman writes:

dissipationless dynamic resistance

Get a patent on that.

Lee DeForest beat me to it by about 100 years.

Gary

On Tue, 6 Jan 2004 10:21:33 -0500, Ned Simmons wrote:
In article ,
says...

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.


All right. Yet another bad electrical analogy. I oughta
know better by now.

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*.

Then please respond to my previous post and explain how
you're going to determine whether the water removed from a
reservoir, or the air released from a pressurized tank, was
used to do mechanical work, or simply released to the
environment. If you're correct it will be possible to tell
without looking beyond the boundaries of the reservoir or
tank.

If I'm correct, and I am, you have to *look* downstream to
see whether the water or air released was usefully employed.
There's no way to determine that from inside the tank.

Fortunately, a regulator is a feedback device, it does "look"
downstream to see what the load demand is, and adjusts
itself to just supply that demand. In other words, it looks
at the downstream pressure and adjusts the flow just enough
to maintain the pressure at the set value. It lets no more air
through than is required to supply the energy the load
demands.

Gary

On Tue, 6 Jan 2004 15:01:54 -0600, "Tim Williams" wrote:
"Gary Coffman" wrote in message
.. .
dissipationless dynamic resistance of a vacuum tube,

So why do my plates glow when they are overbiased? Do you recommend
I grab onto a 6L6 for more than 30 seconds?

Tim

I can't lay my hands on my receiving tube manual at the moment,
but I do have an RCA transmitting tube manual at hand so I'll use
the 813 as my example.

It has a filament voltage of 10 volts at 5 amps. So this converts
electrical energy to heat at the rate of 50 watts. That makes the
tube envelope hot, but is not part of the tube's dynamic resistance.
It is a parasitic loss, necessary for a vacuum tube's operation, but
not part of the energy flow it is controlling. Note that a transistor
wouldn't have this loss, nor does an air regulator.

Static plate voltage is 2250 volts, and static plate current is 50 mA.
So the static plate resistance is 2250/0.05 = 45,000 ohms, and
energy is dissipated at the plate at a rate of 112.5 watts. This is
due to the kinetic impact of electrons on the plate. Again, this is
necessary to the tube's operation, but isn't part of the energy it is
controlling. A transistor would have a similar static collector current
and collector dissipation, but an air regulator would not.

The dynamic resistance is 8,800 ohms, and the energy controlled,
but not dissipated, by that dynamic resistance is at a rate of 573.75
watts. None of that is consumed by the tube. The dynamic resistance
of a tube is the slope of the load line drawn on the characteristic
curves of the tube. It exists because of the throttling effect of grid
voltage on plate current. An air regulator has a similar load line,
or lossless dynamic resistance, controlled by the feedback to the
regulator diaphragm and pintle valve.

Gary

On Tue, 06 Jan 2004 04:04:40 -0600, Richard J Kinch wrote:
Gary Coffman writes:

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.

You are correct that with-the-regulator beats without-the-regulator, but
compressing-the-air-to-50-psi-to-start-with beats them both, which is to
say, if mechanical efficiency is the figure of merit, then a regulator on a
high-pressure supply is wasteful compared to an unregulated source at the
application pressure.

It isn't the regulator which is wasteful. The energy loss is in the compressor
you're asking to pump to a higher tank pressure. And no surprise, it is the
compressor which gets hot as it does so. *That's* where the energy loss in
your scenario occurs, not in the regulator. The regulator itself has nearly
zero loss, as my example of drawing down a filled tank to supply a load
shows.

Gary

In article , Gary Coffman says...

Get a patent on that.

Lee DeForest beat me to it by about 100 years.

LOL

Almost exactly 100 years.

Not that *he* understood anything about that. I take
it you've seen or read "empire of the air?"

Jim

==================================================
please reply to:
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==================================================

On Tue, 06 Jan 2004 03:42:57 -0600, Richard J Kinch
wrote:

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.

We have some HOA customers with a meter pedestal and a meter feeding
a 4-zone irrigation timer - figure a watt or two to run the timer, and
2 or 3 watts when there's a solenoid valve open. 30 years later, and
the watthour meter reads 0000014.

Watt-hour meters can record infintessimally small currents like that
over the long term, but you can't see it to use the information in the
short term. If the disk turns 1 degree...

-- Bruce --
--
Bruce L. Bergman, Woodland Hills (Los Angeles) CA - Desktop
Electrician for Westend Electric - CA726700
5737 Kanan Rd. #359, Agoura CA 91301 (818) 889-9545
Spamtrapped address: Remove the python and the invalid, and use a net.

Don Foreman writes:

What waste? Heat is still energy. If the downstream air is colder
than ambient then heat is not lost by conduction or radiation to
ambient, so that energy is still in the downstream gas.

The point of the dispute was whether overcompressing and then regulating
back down to application pressure was more/less/equally efficient
compared to just compressing to application pressure (or just above).
In short, are regulators inherently "lossy".

Yes, air regulators are inherently lossy. The answer is, all other
things being equal, overcompressing/regulating-down is always *less*
efficient, in a degree increasing with the degree of overcompression and
down-regulating.

This is not to say that practical considerations won't make the
inefficiency necessary. For example, you may have only one supply pipe
across a building going to a variety of tools, and you therefore need to
transmit the pressure for the highest-pressure tool. Or your pipes may
be too small to transmit enough air at lower pressure. Or you may need
very close regulation of the application pressure and widely varying
delivery volumes.

If I may risk another analogy, compressing and transmitting air at
excess pressure (above what is needed for the application) is *not* an
efficiency measure like boosting voltage up and down for long-distance
electric power transmission.

Maybe we're getting closer to understanding, though. We've agreed
that a certain amount (mass) of air is compressed, and some heat is
lost in that process. Compression to higher pressure results in more
work being converted to heat (usually lost) at the compressor or in
the reservoir as that air cools by conduction to ambient -- and
perhaps a little radiation as well. But what about after that, as
in from just before the regulator to point of exhaust? I think that
was the original question.

Yes, that was the original question. But the answer is, regulators are
lossy, even if you assume the compressor is 100 percent efficient and
lossless.

Gary Coffman writes:

It exists because of the throttling effect of grid
voltage on plate current. An air regulator has a similar load line,
or lossless dynamic resistance, controlled by the feedback to the
regulator diaphragm and pintle valve.

I know what you mean by dynamic resistance (at 47 today, mine was the last
electrical engineering class in 1975 to study vacuum tube circuit design).
But I don't follow the lossless.

Forget the vacuum tube details like filament power. Assume they are
perfect valves. Electronic charge is conserved in the device, so the
current-in equals current-out. But voltage-out is less than voltage-in
(definition of a regulator or valve). Thus power-out (voltage times
current) is less than power-in, and the difference is the inherent loss.

"Resistance" is force, ergo work, ergo loss. Doesn't matter if it is
"dynamic" or not, or if it is charge or air flowing.

Gary Coffman writes:

It isn't the regulator which is wasteful. The energy loss is in the
compressor you're asking to pump to a higher tank pressure. And no
surprise, it is the compressor which gets hot as it does so. *That's*
where the energy loss in your scenario occurs, not in the regulator.

No. The compressor can be assumed perfect.

THE MAXIMUM ATTAINABLE WORK DOWNSTREAM OF A RESTRICTION IS NECESSARILY
LESS THAN UPSTREAM OF THAT RESTRICTION.

The operating principle of a conventional regulator is a restriction.
It is just a valve with a feedback arrangement on the handle
Restrictions are lossy. Otherwise we'd be using the thinnest possible
air hoses instead of paying for big ones.

The regulator itself has nearly zero loss, as my example of drawing
down a filled tank to supply a load shows.

I've lost track of who is exhibiting what. But this waving of hands
with Boyle's law and energy being P*V is flawed analysis. It shouldn't
even take analysis. It should be obvious, anything that impedes the
flow of compressed air has got to be robbing power.

In article , Richard J Kinch
says...

It should be obvious, anything that impedes the
flow of compressed air has got to be robbing power.

This is fundamentally incorrect. An orifice that
restricts flow does not disipate power. Consider
an expansion valve in a refrigeration system.

Your approach also fails at one end point, where
the restriction prevents only a tiny bit of flow,
almost none. YOu sould say this is the *lossiest*
case - but it isn't.

If you insist on approaching the issue with an
incorrect view like this, you would never be able
to model the real world behavior of these devices.

Jim

==================================================
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==================================================

In article , Richard J Kinch
says...

Yes, that was the original question. But the answer is, regulators are
lossy, even if you assume the compressor is 100 percent efficient and
lossless.

The process you describe (compressing to a higher pressure
and then regulating down to the point of use) may be less
efficient, overall. But the statement 'regulators are lossy'
is incomplete and inaccurate in light of the overall system
you describe.

There is no dissipation in the regulator itself.

Jim

==================================================
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==================================================

jim rozen writes:

An orifice that restricts flow does not disipate power.

You make a clear statement of what is obviously false, for any plain
meaning of "dissipate power".

The maximum attainable work from a compressed air source, being drawn
through a close restriction, is less than the maximum work attainable drawn
through a more open restriction. That difference is the loss of power
directly attributable to the difference in restriction. This is true
whether one presumes isothermal or adiabadic conditions, or for any ambient
pressure, etc.

It seems hopeless to correct your misunderstanding.

Nevertheless, I presume you try not to pinch your air hoses, even though
you believe restricting flow doesn't "dissipate power".

jim rozen writes:

There is no dissipation in the regulator itself.

I hope this isn't a quibble over what "dissipation" means. There is a loss
of power attributable to the regulator, as a measure of the input versus
the output pressure and mass flow. The body of the regulator itself need
not gain the waste heat.

Richard J Kinch wrote:
Gary Coffman writes:

It isn't the regulator which is wasteful. The energy loss is in the
compressor you're asking to pump to a higher tank pressure. And no
surprise, it is the compressor which gets hot as it does so. *That's*
where the energy loss in your scenario occurs, not in the regulator.

No. The compressor can be assumed perfect.

THE MAXIMUM ATTAINABLE WORK DOWNSTREAM OF A RESTRICTION IS NECESSARILY
LESS THAN UPSTREAM OF THAT RESTRICTION.

The operating principle of a conventional regulator is a restriction.
It is just a valve with a feedback arrangement on the handle
Restrictions are lossy. Otherwise we'd be using the thinnest possible
air hoses instead of paying for big ones.

The regulator itself has nearly zero loss, as my example of drawing
down a filled tank to supply a load shows.

I've lost track of who is exhibiting what. But this waving of hands
with Boyle's law and energy being P*V is flawed analysis. It
shouldn't even take analysis. It should be obvious, anything that
impedes the flow of compressed air has got to be robbing power.

The velocity and turbulence downstream of the orifice is essentially wasted.
That's a lot of kinetic energy that is doing no useful work. However, all
this discussion of where the heat/energy is lost has little to do with using
air to do mechanical work. Compressor inefficiency is also not really the
issue. If we set Pressure*Volume equal to some constant, it takes more work,
even with an ideal compressor, as P goes up and V goes down. A system that
reverses that downstream wastes all of that extra work. I'm saying work
because even an ideal compressor piston will have to exert a greater force
when filling a tank that is at a higher pressure. That is not waste heat or
bypassed air, just an unavoidable fact.

In article , Richard J Kinch
says...

I hope this isn't a quibble over what "dissipation" means.

It's not a quibble. This is not a simple question, and
gaining a good understanding of the principles in operation
require rigorous use of terms.

Jim

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==================================================

In article , Richard J Kinch
says...

Nevertheless, I presume you try not to pinch your air hoses, even though
you believe restricting flow doesn't "dissipate power".

Read the previous comments about series regulators. Then go
back and read them again. The reason that one employs hoses
or pipes large enough to carry the flow is *not* to avoid
thermal losses. It's to ensure that full pressure is available
at the point of use.

To reiterate what has been said a dozen times befo

Any energy not used remains in the reservoir, behind
the restriction.

Jim

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==================================================

I have a bunch of tube manuals in three or four arms reach -

I suspect the glowing plates on the 6L6 is a Beam tube.
I suspect the Screen voltage is a bit high or the Grid voltage
is a bit in the positive direction (should be negative DC and modulate that.)
If the Grid is near zero and you drive over into the + side - the Grid
pre-accelerates boiling electrons gathering on the hot cathode - IIRC thorium enhanced
book still in shelf.

When the grid pre-accelerates - the screen post accelerates as it should, but at a higher
current density now - due to the assist. The plate physically gets smacked with electrons.
Normal amount and the plates can keep their cool. To much and they get hot or melt a hole.

I don't recommend anyone touch any hot tube if it is half the length of a cig lighter.
More than that, and the wattage can far exceed the capacity of your skin to self cool.
a.k.a. burn.

I myself use 5881's two matched pair to drive my Mono output transformer. :-)

Martin

Gary Coffman wrote:
On Tue, 6 Jan 2004 15:01:54 -0600, "Tim Williams" wrote:

"Gary Coffman" wrote in message
. ..

dissipationless dynamic resistance of a vacuum tube,

So why do my plates glow when they are overbiased? Do you recommend
I grab onto a 6L6 for more than 30 seconds?

Tim


I can't lay my hands on my receiving tube manual at the moment,
but I do have an RCA transmitting tube manual at hand so I'll use
the 813 as my example.

It has a filament voltage of 10 volts at 5 amps. So this converts
electrical energy to heat at the rate of 50 watts. That makes the
tube envelope hot, but is not part of the tube's dynamic resistance.
It is a parasitic loss, necessary for a vacuum tube's operation, but
not part of the energy flow it is controlling. Note that a transistor
wouldn't have this loss, nor does an air regulator.

Static plate voltage is 2250 volts, and static plate current is 50 mA.
So the static plate resistance is 2250/0.05 = 45,000 ohms, and
energy is dissipated at the plate at a rate of 112.5 watts. This is
due to the kinetic impact of electrons on the plate. Again, this is
necessary to the tube's operation, but isn't part of the energy it is
controlling. A transistor would have a similar static collector current
and collector dissipation, but an air regulator would not.

The dynamic resistance is 8,800 ohms, and the energy controlled,
but not dissipated, by that dynamic resistance is at a rate of 573.75
watts. None of that is consumed by the tube. The dynamic resistance
of a tube is the slope of the load line drawn on the characteristic
curves of the tube. It exists because of the throttling effect of grid
voltage on plate current. An air regulator has a similar load line,
or lossless dynamic resistance, controlled by the feedback to the
regulator diaphragm and pintle valve.

Gary


--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer
NRA LOH, NRA Life
NRA Second Amendment Task Force Charter Founder

"Martin H. Eastburn" wrote in message
...
If the Grid is near zero and you drive over into the + side - the Grid
pre-accelerates boiling electrons gathering on the hot cathode - IIRC
thorium enhanced
book still in shelf.

811 is an example of a directly heated thoriated tungsten cathode; 6L6 uses
an indirectly heated unipotential oxide coated cathode. Also, red plates
are simply too much dissipation, on a 6L6 at say, 400V screen and plate,
you'll get glowing around oh, -20Vg.

I don't recommend anyone touch any hot tube if it is half the length of a
cig lighter.
More than that, and the wattage can far exceed the capacity of your skin
to self cool.
a.k.a. burn.

And that rule of thumb (and fingers, and hands, and anything else that might
contact a hot tube doesn't apply at all to Hept'AU7:
http://webpages.charter.net/dawill/I...tAU7_Front.jpg
For all that size, weight and heat it only puts out 7W. But that's plenty
(unless I'm listening to Metallica, anyway). :-)

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

jim rozen writes:

To reiterate what has been said a dozen times befo

Any energy not used remains in the reservoir, behind
the restriction.

Let me show the absurdity of this, one more time.

Say we operate a compressed air system at some specific requirement of a
given mass flow and pressure (forget the "cfm" canard for now and just
specify it mass flow). The mass flow of air is the same everywhere in
the output circuit, starting with what leaves the compressor/reservoir,
which then goes into the regulator, and then out of the regulator. The
pressure drops across the regulator, the volume of a unit of gas mass
expands, but the mass flow is still the same everywhere. This is just
conservation of mass, aka Kirchoff's law.

You say the reservoir can be kept at an excess pressure, then regulated
down, and the energy output from the reservoir is the same in either
case, the difference "not used remains in the reservoir".

But the *mass flow* at the reservoir is *equal* in either case, while
the reservoir output pressure is different. Thus the energy flow out of
the reservoir is higher with the regulator, and lower without. The
difference is being wasted by the regulator. This difference does not
"stay in the reservoir", it has gone into the output.

You don't even need to know any equations for how energy relates to mass
and pressure, just that energy in compressed air is an increasing
function of both mass and pressure. This should all be clear enough with
simple mechanical intuition. No need for quantitative thermodynamic
analysis.

In article , Richard J Kinch
says...

No need for quantitative thermodynamic
analysis.

As I said before, you cannot understand compressible gas
flow in systems, using electrical analogies. I will
go out on a further limb and suggest that one cannot
understand the same subject, with a mechanical analogy
either. It's not the same. It doesn't *work*. This
is why thermodynamics was invented.

Jim

==================================================
please reply to:
JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com
==================================================

jim rozen wrote:

As I said before, you cannot understand compressible gas
flow in systems, using electrical analogies.

Hi Jim,

Mass flow isn't an analogy - its the real thing, and conservation of
mass is real. Mass into the system must equal mass out. There is no
way around that. The rocket engine folks use mass flow all the time.

Mass flow is normally used with gasses (except of course in compressor
specifications!), volume flow with liquids. Thinking of mass flow as
current (its the same in all parts of a series circuit), and pressure
as voltage, the electrical analogy works reasonably well (although the
ideal fluid resistor is nonlinear).

Fitch

On Fri, 09 Jan 2004 16:24:31 -0600, Richard J Kinch
wrote:

Don Foreman writes:

What waste? Heat is still energy. If the downstream air is colder
than ambient then heat is not lost by conduction or radiation to
ambient, so that energy is still in the downstream gas.

The point of the dispute was whether overcompressing and then regulating
back down to application pressure was more/less/equally efficient
compared to just compressing to application pressure (or just above).
In short, are regulators inherently "lossy".

Yes, air regulators are inherently lossy. The answer is, all other
things being equal, overcompressing/regulating-down is always *less*
efficient, in a degree increasing with the degree of overcompression and
down-regulating.

This is not to say that practical considerations won't make the
inefficiency necessary. For example, you may have only one supply pipe
across a building going to a variety of tools, and you therefore need to
transmit the pressure for the highest-pressure tool. Or your pipes may
be too small to transmit enough air at lower pressure. Or you may need
very close regulation of the application pressure and widely varying
delivery volumes.

If I may risk another analogy, compressing and transmitting air at
excess pressure (above what is needed for the application) is *not* an
efficiency measure like boosting voltage up and down for long-distance
electric power transmission.

Maybe we're getting closer to understanding, though. We've agreed
that a certain amount (mass) of air is compressed, and some heat is
lost in that process. Compression to higher pressure results in more
work being converted to heat (usually lost) at the compressor or in
the reservoir as that air cools by conduction to ambient -- and
perhaps a little radiation as well. But what about after that, as
in from just before the regulator to point of exhaust? I think that
was the original question.

Yes, that was the original question. But the answer is, regulators are
lossy, even if you assume the compressor is 100 percent efficient and
lossless.

Answer offered is an assertion still dodging the question of where
claimed lost energy might go. This is not a response worthy of an
annointed scholar.

"Don Foreman" wrote in message
news
Answer offered is an assertion still dodging the question of where
claimed lost energy might go. This is not a response worthy of an
annointed scholar.

Well certainly you have power (err...that's not energy...so just go
integrate will ya?) dissipated in the compressor as it compresses the air,
and you likewise have a cooling at the regulator... the power lost in
turbulence (and moving the whatever downstream from it, that is, any
mechanical work, such as a tool, or blowing along more air if it's an open
leak) is where the actual power flows from input (mechanical compression) to
output.

That seems obvious enough to me.

Tim

--
"That's for the courts to decide." - Homer Simpson
Website @ http://webpages.charter.net/dawill/tmoranwms

jim rozen writes:

No need for quantitative thermodynamic
analysis.

As I said before, you cannot understand compressible gas
flow in systems, using electrical analogies. I will
go out on a further limb and suggest that one cannot
understand the same subject, with a mechanical analogy
either. It's not the same. It doesn't *work*. This
is why thermodynamics was invented.

My simple qualitative analysis (consisting of mass conservation plus energy
being an increasing function of increasing mass and/or pressure) fully
disproves your assertions. Thermodynamics is fine, I can do thermo, but it
is more than needed, difficult to display in ASCII, and elaborate enough to
provide perpertual-motion theorists (like you are starting to sound like)
with endless quibbling points.

Don Foreman writes:

Yes, that was the original question. But the answer is, regulators are
lossy, even if you assume the compressor is 100 percent efficient and
lossless.

Answer offered is an assertion still dodging the question of where
claimed lost energy might go.

The "assertion" I have fully analyzed several times now. I am not gonna
repeat that.

No dodging. I offered four destinations for the missing energy: friction
at the orifice, sonic noise, turbulence downstream, work done on the
atmosphere by the expansion. Depending on the regulator design, the
proportion of each will vary.