Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work.

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #241   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Richard J Kinch
says...

... elaborate enough to
provide perpertual-motion theorists (like you are starting to sound like)
with endless quibbling points.


I suggest you go back and re-read the entire thread
and find any other reference to perpetual motion.
If you think that a full understanding of a topic
like this which by it's nature has some math and
equations in it, is quibbling, then I wish you luck
sir.

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

  #242   Report Post  
Don Foreman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Sun, 11 Jan 2004 03:13:48 -0600, Richard J Kinch
wrote:

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.


Those are guesses based on the notion that energy is lost, must go
somewhere so these might be possibilities. They're not plausible.

Friction produces heat but the regulator does not get hot -- so that
energy must remain in the downstream air. Turbulence downstream is ,
uh, downstream, and still contained within the airflow -- doesn't
leave the system unless the lines get warm and they don't. There is
no work done on the atmosphere until the air leaves the system at
point of use. Sonic noise is reallly a reach: the acoustic
threshold of pain (120 dB) is about 12 watts (.016 HP) from a
distance of 1 meter.

I still assert that little or no energy leaves the system at the
regulator, and in fact some may be added if the regulator is colder
than ambient. However, not all energy is available energy.
Available energy derives from the ability of pressure to exert force
and therefore do work if there is motion (expansion). If useful work
is not extracted by a piston, turbine or whatever during expansion,
then it must become heat in the gas even though the temperature drops.
The entropy increases.

That heat is then carried along with the lower pressure gas -- but a
rule of thermodynamics is that heat energy in a gas whose temperature
is below T0, the temperature of the zero-reference "sink" which in
this case is ambient atmosphere at the point of use, is not available
energy that can do work. I therefore submit that available energy
lost in the regulator is converted to heat (albeit at lower
temperature because there has been expansion). This heat energy
remains in the lower-pressure air (is not lost at the regulator
because the regulator runs cold) but since the downstream mass flow
of air is at both lower pressure and lower temperature some of the
energy has become unavailable to do work and leaves the system as heat
content (albeit at low temperature due to expansion) in the exhaust
air.

By the way, Kirchoff's law deals with current in a circuit, has
nothing to do with mass flow.



  #243   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Fitch R. Williams
says...

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).


Absolutely true, but the question was, can one undersand
how energy is transferred in various processes with compressible
fluids, using *only* mass flow considerations?

I would maintain that one has to consider a number of other
issues, like the temperature of the gas at various points
in the cycle, the pressure, the flow velocities, and the amount
of mechanical work it might be performing.

Jim

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

  #244   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Don Foreman says...

By the way, Kirchoff's law deals with current in a circuit, has
nothing to do with mass flow.


Stop, Don. He really wants to view this as bulbs-n-batteries.

JIm

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

  #245   Report Post  
Tim Williams
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

"jim rozen" wrote in message
...
If you think that a full understanding of a topic
like this which by it's nature has some math and
equations in it, is quibbling, then I wish you luck
sir.


Oh Jim, how little you know... it's quibbling but on an RCM scale ;-)

Tim

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




  #246   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

jim rozen writes:

I suggest you go back and re-read the entire thread
and find any other reference to perpetual motion.


You're the perpetual motion advocate. You assert that an expanded quantity
of air represents the same energy as the compressed equal quantity out of a
reservoir, the difference "remains in the tank". But the compressed form
can be made into the expanded, with a yielding of more energy, which is by
defintion a perpetual motion device.
  #247   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Don Foreman writes:

There is
no work done on the atmosphere until the air leaves the system at
point of use.


If it is expanding out of a regulator, the volume must be displaced
*somewhere*, either in the "system" or out of it or some of both. If
the "system" includes, say, another expanding reservoir (an expanding
air cylinder or balloon), then the work is done on the atmosphere
although the air is still in the "system".

If the "system" ends in an ordinary tool with an exhaust port, then,
yes, the extra expansion energy is not all disspated as heat but partly
as work against the atmosphere (which globally just got bigger by that
much volume).

I therefore submit that available energy
lost in the regulator is converted to heat (albeit at lower
temperature because there has been expansion).


Uh, that was my point. Friction, turbulence, sonic noise, all end up as
heat. In theory you could have a perfect regulator that emitted no
noise or turbulence, that worked entirely on friction, something like a
perfectly muffled turbine air motor coupled to a brake.

Sonic noise is reallly a reach


Yes, noise represents a tiny amount of energy, but given the
disputations of this thread, someone would be complaining about it if it
isn't included.

I suppose you could build a regulator out of a siren that would maximize
the proportion of regulator loss emitted as sonic noise.

By the way, Kirchoff's law deals with current in a circuit, has
nothing to do with mass flow.


Kirchoff's law as reapplied to air currents instead of electron
currents, then. I assumed that would be clear that I didn't mean it
literally. The same principle applies, conservation of mass flow,
whether gas molecules or electrons.
  #248   Report Post  
Don Foreman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On 11 Jan 2004 09:51:51 -0800, jim rozen
wrote:
Stop, Don. He really wants to view this as bulbs-n-batteries.

JIm


OK, I will.
  #249   Report Post  
Don Foreman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Mon, 12 Jan 2004 00:08:32 -0600, Richard J Kinch
wrote:

Don Foreman writes:

There is
no work done on the atmosphere until the air leaves the system at
point of use.


If it is expanding out of a regulator, the volume must be displaced
*somewhere*, either in the "system" or out of it or some of both. If
the "system" includes, say, another expanding reservoir (an expanding
air cylinder or balloon), then the work is done on the atmosphere
although the air is still in the "system".

If the "system" ends in an ordinary tool with an exhaust port, then,
yes, the extra expansion energy is not all disspated as heat but partly
as work against the atmosphere (which globally just got bigger by that
much volume).


That works the same with or without a regulator.

I therefore submit that available energy
lost in the regulator is converted to heat (albeit at lower
temperature because there has been expansion).


Uh, that was my point. Friction, turbulence, sonic noise, all end up as
heat. In theory you could have a perfect regulator that emitted no
noise or turbulence, that worked entirely on friction, something like a
perfectly muffled turbine air motor coupled to a brake.


The brake would get hot and dissipate to ambient while the regulator
does not. A regulator does work almost entirely on viscous friction,
The question continues to be where the energy goes since the regulator
does not get not. To get it right you must deal with entropy and
available energy. Any irreversale process, by definition, involves
an increase in entropy. Expansion of air thru an orifice as in a
regulator is definitely an irreversable process so entropy is
increased. That's where the energy goes: out the exhaust port as
heat energy unavailable for doing useful work. It is NOT lost in
the regulator, total energy is preserved, but the regulator does
increase entropy and therefore reduces available energy downstream.

Bottom line is less power to the tool. I just wanted to understand
why. I think I do now so I'll shut up. I've learned a lot here.



  #250   Report Post  
ATP
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Don Foreman wrote:
On Mon, 12 Jan 2004 00:08:32 -0600, Richard J Kinch
wrote:

Don Foreman writes:

There is
no work done on the atmosphere until the air leaves the system at
point of use.


If it is expanding out of a regulator, the volume must be displaced
*somewhere*, either in the "system" or out of it or some of both. If
the "system" includes, say, another expanding reservoir (an expanding
air cylinder or balloon), then the work is done on the atmosphere
although the air is still in the "system".

If the "system" ends in an ordinary tool with an exhaust port, then,
yes, the extra expansion energy is not all disspated as heat but
partly as work against the atmosphere (which globally just got
bigger by that much volume).


That works the same with or without a regulator.

I therefore submit that available energy
lost in the regulator is converted to heat (albeit at lower
temperature because there has been expansion).


Uh, that was my point. Friction, turbulence, sonic noise, all end
up as heat. In theory you could have a perfect regulator that
emitted no noise or turbulence, that worked entirely on friction,
something like a perfectly muffled turbine air motor coupled to a
brake.


The brake would get hot and dissipate to ambient while the regulator
does not. A regulator does work almost entirely on viscous friction,
The question continues to be where the energy goes since the regulator
does not get not. To get it right you must deal with entropy and
available energy. Any irreversale process, by definition, involves
an increase in entropy. Expansion of air thru an orifice as in a
regulator is definitely an irreversable process so entropy is
increased. That's where the energy goes: out the exhaust port as
heat energy unavailable for doing useful work. It is NOT lost in
the regulator, total energy is preserved, but the regulator does
increase entropy and therefore reduces available energy downstream.

Bottom line is less power to the tool. I just wanted to understand
why. I think I do now so I'll shut up. I've learned a lot here.


It's been a good thread. I think there are still a few holdouts as far as
accepting your last paragraph:-)




  #252   Report Post  
Ned Simmons
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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


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.


So what you're saying is that you can release a pound of
air from a tank pressurized to 100psi and either:

A. Feed the air directly into a cylinder and do x ft*lbs of
mechanical work.

or

B. Run the released air thru a regulator set to 50 psi and
do a smaller amount of work.

If what you say is true, the difference in the amount of
mechanical work between the two cases will "remain in the
tank", but you can't detect that difference in energy
without looking outside the tank. You don't see the
contradiction here?


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.


The regulator does not look downstream at the load demand,
it sees only the pressure at its outlet and adjusts the
flow to maintain this pressure. This in no way guarantees
optimal efficiency in the use of the energy available in
the air.

Ned Simmons
  #253   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Ned Simmons
says...

[his] 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.


Unless you're treating the tiny orifice as a leak, it is
indeed the "lossiest" case. For a given flow, greater delta
P across the orifice results in a greater loss.


My 'endpoint' was where the flow though the leak
was zero. No flow at all.

Jim

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

  #254   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Richard J Kinch
says...

You're the perpetual motion advocate. You assert that an expanded quantity
of air represents the same energy as the compressed equal quantity out of a
reservoir,


Incorrect.

I said that the total energy of the system remains a constant,
once one accounts for the energy in the following forms:

1) internal (thermal) energy of all the gas

2) stored energy in the various reservoirs

3) kinetic energy in the gas flows

4) mechanical work done by, or on, the gas

5) heat loss or gain of the gas to or from its surroundings.

You need to be able to believe in conservation of energy to
buy into all this, but I assure you that if the accounting
is done properly, it all works out.

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

  #255   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Don Foreman says...

... To get it right you must deal with entropy and
available energy.


Yep. But he wants to avoid thermodynamics. I think
it's an honest quandry.

Jim

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



  #256   Report Post  
Gary Coffman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Mon, 12 Jan 2004 08:46:21 -0500, Ned Simmons wrote:
In article ,
says...
On Tue, 6 Jan 2004 10:21:33 -0500, Ned Simmons wrote:
In article ,
says...


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.


So what you're saying is that you can release a pound of
air from a tank pressurized to 100psi and either:

A. Feed the air directly into a cylinder and do x ft*lbs of
mechanical work.

or

B. Run the released air thru a regulator set to 50 psi and
do a smaller amount of work.


To get down to 50 PSI on the load side of the regulator,
you either have to release half as much air as you would
to have 100 PSI downstream of the regulator, or give the
air twice the volume to expand in at the load.

P is proportional to n/V.

So if you release half as much air to get 50 PSI downstream
of the regulator, then the remainder does stay in the tank. If
you release the same mass, then you need a cylinder with twice
the piston area for it to expand against. In either case, all the
work is accounted for, and it is roughly speaking the same
amount of work.

If what you say is true, the difference in the amount of
mechanical work between the two cases will "remain in the
tank", but you can't detect that difference in energy
without looking outside the tank. You don't see the
contradiction here?


You can't measure work if no work is done. How much
work is done by the air sitting in the tank? None. And
as an aside, power is defined as work done per unit time,
P = W/t, so if work is zero in the tank, so is power.

You have to look at the load to see any work done, and
that's outside the tank. For example if the load is an
air cylinder, you can measure how far it moves a weight,
W=F*d, and then you can measure how long that took,
and have a value for power, P=W/t.

You can measure changes in the *potential* energy in
the tank by measuring changes in the mass and pressure
of the air in the tank, but you can't tell anything about work
or power without looking at the load outside 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.


The regulator does not look downstream at the load demand,
it sees only the pressure at its outlet and adjusts the
flow to maintain this pressure. This in no way guarantees
optimal efficiency in the use of the energy available in
the air.


You have to assume that the regulator set pressure was
chosen by the system designer to match the pressure demand
of the load. Just as computers can't actually think, neither can
regulators. That's the system programmer or designer's job.
The computer or regulator's job is merely to follow instructions.
In the case of a series regulator, it varies the amount of air
released in order to maintain the set pressure against load
demand. The air it doesn't release *remains in the tank*.

Gary
  #257   Report Post  
Ted Edwards
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

There's a statement in one of Heinlein's stories that goes
(approximately) as, "You can't explain how television really works to
Grandma. She just doesn't have the math." I've been sort of following
pieces of this thread and think this Kinch guy is like that - no
comprehension of Integral P dV and no wish to learn anything new.

Maybe it's time we left him to his religion.

Ted

  #258   Report Post  
Grant Erwin
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Ted Edwards wrote:

There's a statement in one of Heinlein's stories that goes
(approximately) as, "You can't explain how television really works to
Grandma. She just doesn't have the math." I've been sort of following
pieces of this thread and think this Kinch guy is like that - no
comprehension of Integral P dV and no wish to learn anything new.


Well, I'll give him this much. He passionately believes what he is saying
even if he isn't much good at saying it. Personally, I don't believe I
truly understand something until I can explain it simply. I started this
thread simply wondering if you could get more volume out of a regulator
than what went into it, and this thread has really diverged from that.

I explicitly phrased my question to state that the only air flow I was
interested in was what went through the regulator, although many many
people took it to mean I wanted to understand how an air compressor works
or what is efficient or inefficient about it. I didn't care in the
slightest about that. If I had it to do over again I would phrase the
question with an imaginary sea god blowing into a tube such that
10 CFM @ 180 psi @ some temp were flowing down the tube, and ask if
that tube went into a regulator and came out at the same temp at 90 psi
how many CFM would be flowing. Maybe then people would "get" that I wasn't
asking about air compressors.

It was very interesting to see the range of answers to the above question.
Some people thought you only get 10 CFM coming out. Others thought you get
20 CFM out. I then realized the problem isn't specified correctly because
I didn't say whether it was 180 psi ABSOLUTE or 180 psi GAUGE. I learned
quite a bit in that area, and hopefully some others did too.

I haven't followed this thread recently because it has degenerated into
technobabble and a lot more heat than light. I know Ted Edwards doesn't
play that game, so I thought I'd read what he wrote. So maybe this can
act like sort of a summary and maybe the remaining combatants would kindly
(and appropriately) start a new thread so I don't have to feel guilty for
inflicting this on all of you anymore.

Grant Erwin
Kirkland, Washington

  #259   Report Post  
Ned Simmons
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article ,
says...
On Mon, 12 Jan 2004 08:46:21 -0500, Ned Simmons wrote:
In article ,
says...
On Tue, 6 Jan 2004 10:21:33 -0500, Ned Simmons wrote:
In article ,
says...


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.


So what you're saying is that you can release a pound of
air from a tank pressurized to 100psi and either:

A. Feed the air directly into a cylinder and do x ft*lbs of
mechanical work.

or

B. Run the released air thru a regulator set to 50 psi and
do a smaller amount of work.


To get down to 50 PSI on the load side of the regulator,
you either have to release half as much air as you would
to have 100 PSI downstream of the regulator, or give the
air twice the volume to expand in at the load.

P is proportional to n/V.

So if you release half as much air to get 50 PSI downstream
of the regulator, then the remainder does stay in the tank. If
you release the same mass, then you need a cylinder with twice
the piston area for it to expand against. In either case, all the
work is accounted for, and it is roughly speaking the same
amount of work.


Assume you release equal masses of air, the pistons have
the same area, and the cylinders are long enough to fully
expand the air. The 100 psi (assume absolute pressure)
piston will exert twice the initial force and will also
have travelled over twice as far when the air is fully
expanded. The force/displacement curve is not linear (Don
already caught me on that), but it should be obvious that
the total work available from the piston is considerably
greater in the unregulated case.


If what you say is true, the difference in the amount of
mechanical work between the two cases will "remain in the
tank", but you can't detect that difference in energy
without looking outside the tank. You don't see the
contradiction here?


You can't measure work if no work is done. How much
work is done by the air sitting in the tank? None. And
as an aside, power is defined as work done per unit time,
P = W/t, so if work is zero in the tank, so is power.

You have to look at the load to see any work done, and
that's outside the tank.


Yes, that was my point.


You have to assume that the regulator set pressure was
chosen by the system designer to match the pressure demand
of the load. Just as computers can't actually think, neither can
regulators. That's the system programmer or designer's job.
The computer or regulator's job is merely to follow instructions.
In the case of a series regulator, it varies the amount of air
released in order to maintain the set pressure against load
demand. The air it doesn't release *remains in the tank*.


The example above, where the load is always in equilibrium
with the piston, is the optimal isothermal case for
extracting the maximum work from the system. It's not a
practical real world example where the piston would be
expected to move at a realistic speed, and that does
require some intelligence in setting the pressure.
Completely characterizing the tradeoffs would be pretty
complicated. I don't mean to imply that regulators are not
useful and should be avoided, but the example does
demonstrate that losses are a consequence of putting a
regulator in a pneumatic system.

Ned Simmons

  #261   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Don Foreman writes:

That's where the energy goes: out the exhaust port as
heat energy unavailable for doing useful work. It is NOT lost in
the regulator,


Not necessarily. A braked turbine is one way to build a regulator. You
could regulate down and take off some of that work at the same time, and
less heat would go out the exhaust.

I repeat that "lossy" means "attributable to the device", whether or not
the device itself gains the waste heat. This was to refute the original
these of this thread that regulators were not lossy.
  #262   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Ted Edwards writes:

I've been sort of following
pieces of this thread and think this Kinch guy is like that - no
comprehension of Integral P dV and no wish to learn anything new.


The only "new" ideas I've heard in this thread are (in sum) "1: CFM does
not mean free air" and "2: Regulators don't waste power", both of which are
false.

If you don't believe my explanations, or don't accept them on the authority
of my engineering credentials, then there's not much else I can do to
persuade you.
  #263   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Gary Coffman writes:

P is proportional to n/V.


Not for adiabatic conditions, which is what practical compressed air
systems most resemble:

p1 * v1^(5/3) = p2 * v2^(5/3)

not:

p1 * v1 = p2 * v2.

So if you release half as much air to get 50 PSI downstream
of the regulator, then the remainder does stay in the tank.


You confuse the p*v energy with the "maximum attainable work". Not the
same. A reservoir of given volume and pressure is not equivalent to double
that volume at half that pressure. Work is done by the former going to the
latter, or work is required to achieve the former from the latter.
  #264   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

jim rozen writes:

You're the perpetual motion advocate. You assert that an expanded
quantity of air represents the same energy as the compressed equal
quantity out of a reservoir,


Incorrect.


OK, so you don't assert that. You must believe (with me) that "same
energy" above should read "different energy". So you must agree with
this:

(1) An expanded mass of air can provide less attainable work than the
same mass when more compressed.

And thus the following:

(2) A given mass flow at a higher pressure holds less attainable work
flow than the same mass flowing at a lower pressure.

(3) A regulator, since the input and output correspond to the
alternatives of (2), must necessarily introduce a loss of attainable
work in the output versus the input.

(4) Which is to say, regulators are inherently inefficient. They
dissipate some of the available work into (ultimately) waste heat or
work performed on the atmosphere.
  #265   Report Post  
Gunner
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Mon, 12 Jan 2004 19:17:30 GMT, Ted Edwards
wrote:

There's a statement in one of Heinlein's stories that goes
(approximately) as, "You can't explain how television really works to
Grandma. She just doesn't have the math." I've been sort of following
pieces of this thread and think this Kinch guy is like that - no
comprehension of Integral P dV and no wish to learn anything new.

Maybe it's time we left him to his religion.

Ted


TV, like the PC, routers and the intergrated circuit, work by FM.

Gunner



"Aren't cats Libertarian? They just want to be left alone.
I think our dog is a Democrat, as he is always looking for a handout"
Unknown Usnet Poster

Heh, heh, I'm pretty sure my dog is a liberal - he has no balls.
Keyton


  #266   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Richard J Kinch writes:

This was to refute the original these ...


Ooops, "thesis".
  #267   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Richard J Kinch writes:

(2) A given mass flow at a higher pressure holds less attainable work
flow than the same mass flowing at a lower pressure.


Ooops again, that should be, "holds more attainable work".
  #268   Report Post  
Gary Coffman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Mon, 12 Jan 2004 15:24:00 -0800, Grant Erwin wrote:
I explicitly phrased my question to state that the only air flow I was
interested in was what went through the regulator, although many many
people took it to mean I wanted to understand how an air compressor works
or what is efficient or inefficient about it. I didn't care in the
slightest about that. If I had it to do over again I would phrase the
question with an imaginary sea god blowing into a tube such that
10 CFM @ 180 psi @ some temp were flowing down the tube, and ask if
that tube went into a regulator and came out at the same temp at 90 psi
how many CFM would be flowing.


You can't answer that question without knowing something about the
downstream load. A regulator works by releasing only enough air to
cause the downstream pressure to rise to the set point. If the downstream
is a sealed tube, flow stops quickly as pressure rises to 90 PSI and remains
there. If the downstream load consumes air, ie does work with it, then the
flow from the regulator increases above zero flow until the amount of air
flow is just sufficient to make up for the load consumption.

So it doesn't matter how large the air mass is above the regulator, or
how much pressure exists before the regulator, the mass of air which
the regulator allows to flow downstream depends solely on the set
pressure and the load demand. The regulator isn't a resistance, it
is a demand controlled valve with a variable area aperture, capable
of opening or closing completely under control of the diaphram, which
is responding to downstream pressure. If you don't know the demand,
you can't know the downstream flow.

Air and its associated energy not needed to meet the demand simply
remains upstream of the regulator, normally in a tank, but in your
Sea God case, in his lungs. It is emphatically *not* dissipated in the
regulator.

Gary
  #269   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Richard J Kinch
says...

(4) Which is to say, regulators are inherently inefficient. They
dissipate some of the available work into (ultimately) waste heat or
work performed on the atmosphere.


Like gary says, "it's not in the regulator."

The regulator is not dissipating power. The
pair systems you could imagine, one with, and one
without, a pressure regulator, can be more or
less efficient and performing some task with
compressed air.

Jim

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

  #270   Report Post  
Ted Edwards
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Richard J Kinch wrote:

not mean free air" and "2: Regulators don't waste power", both of which are
false.


You refuse to do the math (Thermodynamics) and you refuse to acknowledge
experimental evidemce and you don't believe in conservation of energy.

If you don't believe my explanations, or don't accept them on the authority
of my engineering credentials,


Are your "credentials" in Psychiatry? They certainlt aren't in Physics
or Engineering!

then there's not much else I can do to persuade you.


Correct.

I have a heavy duty (7") air driven grinder/wire brush. When cleaning
up a largish piece of rusty metal, my compressor can't readily keep up.
Periodically I give it a break. When I do, I frequently go over and put
ny hand in various places to see if anything is overhaeting. THE
REGULATOR DOES **NOT** GET HOT. Until you get your simple mind around
that FACT, this is a waste of time.

PLONK

Ted




  #271   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Richard J Kinch
says...

I repeat that "lossy" means "attributable to the device", whether or not
the device itself gains the waste heat.


This is a counter-intuitive definition. Few would
accept it as reasonable. However using that definion
is just about the only way you can get your 'lossy
regulator' to work out in practice.

"System containing regulator is less efficient than
one not containing one" would have to be made
equal to "regulators cause loss."

It's a stretch, to be sure.

Jim

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

  #272   Report Post  
Don Foreman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Mon, 12 Jan 2004 15:24:00 -0800, Grant Erwin
wrote:

I explicitly phrased my question to state that the only air flow I was
interested in was what went through the regulator, although many many
people took it to mean I wanted to understand how an air compressor works
or what is efficient or inefficient about it. I didn't care in the
slightest about that. If I had it to do over again I would phrase the
question with an imaginary sea god blowing into a tube such that
10 CFM @ 180 psi @ some temp were flowing down the tube, and ask if
that tube went into a regulator and came out at the same temp at 90 psi
how many CFM would be flowing. Maybe then people would "get" that I wasn't
asking about air compressors.


When you ask what time it is you'd rather skip the lecture on how the
clock works? That's certainly reasonable enough!

If the CFM you mean are the kind often used in compressed air
parlance, aka SCFM or volume of the gas at standard atmospheric
conditions, CFM out = CFM in regardless of pressure drop or
temperature. A cubic foot of air at standard atmospheric condx is a
certain mass of air, and mass is neither lost nor gained in a
regulator.

If you want absolute CFM, the equations describing compression and
expansion in terms of pressure, volume and temperature for isothermal
or adiabatic expansion (different cases) can be found in Machinery's
Handbook and Marks. The case you cite, same temp downstream and
upstream, is isothermal expansion.

I'm not going to type them, but I'd be glad to scan pp 2369 and 2370
of Machinery's Handbook 23 and send them to you by email. Drop me
an email if you'd like me to do so and I'll do it right away.
  #273   Report Post  
ATP
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

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

I repeat that "lossy" means "attributable to the device", whether or
not the device itself gains the waste heat.


This is a counter-intuitive definition. Few would
accept it as reasonable. However using that definion
is just about the only way you can get your 'lossy
regulator' to work out in practice.

"System containing regulator is less efficient than
one not containing one" would have to be made
equal to "regulators cause loss."

It's a stretch, to be sure.

Jim

Not in the context of this thread, which started with a regulator equals
transformer analogy, which many apparently agreed with initially, at least
to some extent. Richard was fundamentally right in disputing the analogy,
along with Ned Simmons.


  #274   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Gary Coffman writes:

The regulator isn't a resistance ...


What nonsense. It's as much a resistance as you pinching an air hose,
or you partly opening or closing an ordinary manual valve in the line.
The only difference in a regulator is that the opening and closing is
controlled by the downstream conditions, instead of your hand. If your
hand is quick enough, you can read a gage and do *exactly* what a
regulator does.

Air and its associated energy not needed to meet the demand simply
remains upstream of the regulator, normally in a tank, but in your
Sea God case, in his lungs. It is emphatically *not* dissipated in the
regulator.


More nonsense. Only the mass is conserved to remain upstream. The
pressure of the mass of air passing through the regulator drops (the
nature of a regulator, after all), and thus the regulator introduces a
waste of some of the energy available for work at the input condition
versus the output, whether or not you care to label this loss
"dissipation", whether or not the end result is heat or not, whether or
not the regulator itself gets hotter or colder.

It is stupid to believe a restriction (dynamically controlled or not)
transforms pressure and volume from input to output with no significant
loss of work potential.
  #275   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

jim rozen writes:

(4) Which is to say, regulators are inherently inefficient. They
dissipate some of the available work into (ultimately) waste heat or
work performed on the atmosphere.


Like gary says, "it's not in the regulator."


If you mean, the manifestation of the loss is not in the regulator
(instead in waste heat downstream, or work spent outside the system), fine.
The regulator nevertheless introduces and accounts for the loss. It sounds
rather coy and Clintonesque to say it is not "in" the regulator. That's
like whining that mechanical inefficiencies aren't "in" an engine, but
flushed out the exhaust. When you're wrong on the physics, you can always
retreat on the definitions.


  #276   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

Ted Edwards writes:

If you don't believe my explanations, or don't accept them on the
authority of my engineering credentials,


Are your "credentials" in Psychiatry? They certainlt aren't in
Physics or Engineering!


You are completely, ignorantly, insultingly, in error.
  #277   Report Post  
Richard J Kinch
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

jim rozen writes:

I repeat that "lossy" means "attributable to the device", whether or not
the device itself gains the waste heat.


This is a counter-intuitive definition. Few would
accept it as reasonable.


No, it is the only appropriate definition.

You use a hose, you *lose* some power "in the hose". Hoses are lossy. The
longer and skinnier they are, the more lossy they are. The hose doesn't
get hot, though. A regulator is just a short hose that gets skinny or fat.
  #278   Report Post  
jim rozen
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

In article , Richard J Kinch
says...

When you're wrong on the physics, you can always
retreat on the definitions.


Works for you.

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

  #279   Report Post  
Gary Coffman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Fri, 09 Jan 2004 16:46:45 -0600, Richard J Kinch wrote:
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.


Work is not a time function. You can conduct air from a tank to a piston
though either a large pipe, or a restricted pipe. The resulting work done
by the piston is the same, only the rate of doing the work changes.
When doing an energy accounting (change in potential energy vs
work done), rates don't matter.

Air is driven through the aperture of a valve by a force which is the
product of the pressure differential existing across the aperture times
the area of the aperture. A small aperture will necessarily have less
force pushing air through it, and will push less air in a given time. But
that doesn't affect the amount of work the air can do on the load side.

Work isn't a time function. Work is force times distance. How long it
takes is irrelevant. Less work is done pushing air through a small
aperture than through a large one of the same length. In other words,
open the valve fully, no regulation, and more work is required to move
air through the length of the valve than would be required if the valve
were nearly closed.

Of course you move more air per unit time in the former case, but
work isn't a time function, so that's not relevant when calculating
dissipation.

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.


Power is expressed at the point work is done. It is the rate of doing
work. P=W/t. There's no work being done by air in a reservoir, so there
is no power there. There is work being done transporting air to the
load, and work done in the load. The latter is the desired work, the
former the parasitic loss. As shown above, a large bore valve/hose/pipe
requires more work to transport air than a small bore regulator/hose/pipe.
So parasitic loss, energy dissipation, is less with the more restricted
system.

You keep confusing power and energy. They aren't the same thing.
Energy is the capacity to do work, or equivalently for accounting
purposes, the amount of work done. Power is just the rate at which
work is done at some point or points in a system. It is a time function,
a rate, energy and work are not.

Gary
  #280   Report Post  
Gary Coffman
 
Posts: n/a
Default SCFM vs. CFM, also air flow/pressure across a regulator

On Sat, 10 Jan 2004 01:19:26 -0600, Richard J Kinch wrote:
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.


Please do. Show your absurdities.

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.


Now *why* does the pressure drop on the downstream side of the
regulator? If you have the same size pipe on the downstream side,
and the same mass of gas flowing in both the upstream and downstream
pipes, where did the pressure go? (Assume infinitely long pipes in
each direction, and that the steady state has been reached, to
simplify your answer.)

Until you provide a place for the air to expand, P remains proportional
to n/V, so upstream and downstream pressures are equal in the two
equal size pipes. Same mass, same volume, equals same pressure.
The gas laws insist upon it. The only way pressure can be lower
downstream of the regulator is if the volume into which it can expand
is larger than the upstream volume.

In the steady state, that means a bleed to atmosphere somewhere
downstream of the regulator. Without that bleed, there is no pressure
drop downstream of the regulator. If the bleed is *beyond* the load,
as is usually the case, no energy is "dissipated" until the exhaust
volume is reached. *That's* where energy is dissipated, not in the
regulator.

Note that this is also where energy is dissipated *if there is no
regulator present*. The presence or absence of a regulator
doesn't affect *where* the energy is dissipated, only the amount
that is dissipated at any given instant (more without the regulator
than with it, because the regulator is what limits mass flow).

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.


The pressure in the reservior is only decreased in proportion to the
mass flow of air out of it. In other words, P is proportional to n/V and
the V of the tank is fixed. Until you draw down the reservior significantly,
ie lower n in the reservoir significantly, reservoir pressure doesn't change
significantly. So the reservoir output pressure at any given instant is
simply a function of the mass flow out of the reservoir.

(With a compressor pounding more air into the reservoir, the output
pressure is simply a function of the net change in air mass in the tank,
which might be upward or downward at any given instant, depending
on which flow is greater.)

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.


Since the way a regulator works is to *limit* mass flow (n) so that the
ratio n/V (pressure) is a certain set amount smaller downstream than
upstream, the reservoir pressure remains higher *with* the regulator
than with just an open pipe.

Since the potential energy in the reservoir is a function of how tightly
the air "spring" in the tank is wound (for a fixed V, this potential energy
is a function of n), more of the energy remains in the tank at any given
instant when mass flow is limited than when the air is allowed to gush
out through a large unrestricted opening.

The whole purpose of a regulator is to be responsive to load demand.
It allows only enough mass flow to occur to satisfy the load's demand
at the set pressure. Since this mass flow is necessarily smaller than
the flow the reservoir could provide without the restriction, more air,
and hence more potential energy, remains in the reservoir when a
regulator is present than when one is absent.

So the energy not required to meet load demand at the set pressure
does indeed remain in the reservoir. It is not dissipated in the regulator.
It never *reaches* the regulator. It is still in the tank waiting to be allowed
to flow to the load, and ultimately to exhaust to atmosphere where any
remaining energy not usefully converted to work by the load *is* dissipated.

Gary
Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
relief valve or back pressure regulator Glenn Ashmore Metalworking 4 July 26th 03 04:07 AM


All times are GMT +1. The time now is 10:38 AM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 DIYbanter.
The comments are property of their posters.
 

About Us

"It's about DIY & home improvement"