Home |
Search |
Today's Posts |
|
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 | Display Modes |
#241
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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:-) |
#251
|
|||
|
|||
SCFM vs. CFM, also air flow/pressure across a regulator
|
#252
|
|||
|
|||
SCFM vs. CFM, also air flow/pressure across a regulator
|
#253
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
SCFM vs. CFM, also air flow/pressure across a regulator
|
#261
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
SCFM vs. CFM, also air flow/pressure across a regulator
Richard J Kinch writes:
This was to refute the original these ... Ooops, "thesis". |
#267
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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
|
|||
|
|||
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 |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Forum | |||
relief valve or back pressure regulator | Metalworking |