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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. |
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#321
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SCFM vs. CFM, also air flow/pressure across a regulator
Good story John - I had one almost like that at GM where I worked for one summer
before my job started. (GM paid 3x more - nice to get a start like that). My issue was sent in - can't recall what it was but it would be a savings. I got a call by my bosses boss, thought that was to be a good call, but was told mine was an exact copy of one accepted two days earlier. I asked exact ? He said word for word. And I was suspect. There was a rotten apple above me somewhere is all I knew. Martin Engineman1 wrote: Years ago I worked for the 3M Co. in their complex in St. Paul MN. It covered about 3/4 sq. mile and had about 18 buildings. I had just completed a 2 yr. company funded AC&R course. Even before that I was an evangelical supporter of the law of conservation of energy (and matter). The company came out with a program saying that whoever could come up with a program that would save the company money, they would reward the person with a generous amount of the savings. I went over to the boilerhouse and saw 3 - 5000 HP steam turbine driven R-12 compressors which were delivering a huge amount of liquid refrigerant at 175 PSI through an orifice to almost atmospheric. It seemed to me that there must be a lot of energy disappearing. I considered the idea of a metorite entering the earth's atmosphere at high speed. At first it develops heat by friction but the relative motion between the gas and solid produces cooling. Later in the process different factors take over and the meteorite burns up. I surmise that the same thing is happening here but at a lower velocity level and the heat is carried away by the refrigerant which is cool because it has evaporated.My idea was to install some sort of turbine-generator in place of the orfice to extract the energy that would otherwise be wasted as heat. My research told me this is commonly done in air and gas liquefaction plants. I made some calculations and concluded that this could result in a savings of 8 to 10 % in the cost of air conditioning. I submitted my idea but the head honchos said that there were easier ways to get that kind of savings in the system. Too bad They didn't tell me what they were as I would have submitted them. In the end I came away empty-handed. Engineman1 -- Martin Eastburn, Barbara Eastburn @ home at Lion's Lair with our computer NRA LOH, NRA Life NRA Second Amendment Task Force Charter Founder |
#322
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air regulator (was SCFM ..)
On 17 Jan 2004 16:04:45 -0800, jim rozen
wrote: The n above gives the number of atoms, basically. In that equation, the V really is the actual volume. In your example, if you have a cubic foot of a gas at 100 psia, at say, 300K, then I can tell you how many moles of gas there are. OK. That does make sense but I'm still not sure about all of the energy. I'll be near a Borders bookstore next week, willl look for the Fermi book. It sounds like something I need to read. |
#323
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SCFM vs. CFM, also air flow/pressure across a regulator
On Tue, 13 Jan 2004 00:51:21 -0600, Richard J Kinch wrote:
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. Irrelevant. PV=nRT. With RT relatively constant, PV=kn, or P=kn/V, and ignoring k, P is proportional to n/V just as I said. 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". No I don't, but you seem hopelessly lost. 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. Well, duh! What does that have to do with air not released from a reservoir by a regulator? Gary |
#324
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SCFM vs. CFM, also air flow/pressure across a regulator
In article , Bruce L. Bergman
writes: Reality check: R-11 or R-12? They use R-11 in huge centrifugal compressor chiller systems like that, and R-12 in automotive systems. I read that far and went "what's wrong with that statement?..." That was a long time ago but It seems that they were using R-12 because it was cheaper and it operated at a lower temp. which meant tha the heat transfer rate would be greater. Engineman1 |
#325
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air regulator (was SCFM ..)
Lewis Hartswick writes:
The case of venting it into vacuum is the same as venting it to atmosphere. I disagree. Nothing to push against, no force, no work, no heat. Just eternal expansion. I think I see another like my dad. He always said a rocket would not work in space since there was nothing to "push against". :-) Mocking one's father is just wrong, like your physics. |
#326
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air regulator (was SCFM ..)
Norman Yarvin writes:
So suppose you have two containers, one filled with gas, and another empty (filled with vacuum). You connect them, and open the valve between them. The gas molecules bounce around into the newly connected container, so now you have two containers each with half the gas. But the gas temperature is the same (or very close). No, adiabatic expansion, temperature falls. |
#327
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SCFM vs. CFM, also air flow/pressure across a regulator
Gary Coffman writes:
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. Irrelevant. PV=nRT. With RT relatively constant, PV=kn, or P=kn/V, and ignoring k, P is proportional to n/V just as I said. Repeat, "adiabatic", quite relevant. T is not constant. |
#328
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air regulator (was SCFM ..)
In article , Richard J Kinch
says... Mocking one's father is just wrong, like your physics. I don't seem to recall that Lewis asserted any phyisics, except perhaps the fact that rockets work in outer space. As for mocking one's father, well - that's another issue. I think that most fathers want their kids to grow up independent and strong, emotionally. Part of that involves a certain amount of, ahem, 'testing' from the kid to the parent. I think this happens between the ages of 15 and 45 or so. I am beginning to experience this myself as I am currently under inspection for my clothing choice. For example I tried to buy my wife some stuff for christmas and my daughter (who was supervising this shopping) pointed out that I had no clue and to simply purchase what she instructed. This was actually the easist christmas foray I've ever had, I will bring her with every time from now on. My own father and I have a certain give and take as well. He is a well respected professional in his own field, but that field is at skew lines to mine. So very early on it was apparent that he should not be trusted with any kind of engineering or electrical issue. I mentioned before that he really does not truly comprehend how alternating currents 'work' and this is entirely true. He did however have the wherewithall to a) insist his kids complete four years of college, and b) to foot the bill for all of it. In my book that overcomes the first shortcoming. Nonetheless there were a number of incidents that occurred as I was growing up, involving vaporized screwdrivers. Basically my knowledge base on electrical issues was ramping up beyond his own abilities and this caused me considerable unease at the time. Not to mention his attempts at amateur plumbing. He looks back at those times with considerable amusement now, and *he* is the one to tease me about it in reverse a bit, if I happen to be deep in some re-wiring, he will show up and threaten to 'check it out for me,' wielding a large screwdriver.... Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== |
#329
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SCFM vs. CFM, also air flow/pressure across a regulator
Gary Coffman wrote:
On Tue, 13 Jan 2004 00:51:21 -0600, Richard J Kinch wrote: 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. Irrelevant. PV=nRT. With RT relatively constant, PV=kn, or P=kn/V, and ignoring k, P is proportional to n/V just as I said. 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". No I don't, but you seem hopelessly lost. 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. Well, duh! What does that have to do with air not released from a reservoir by a regulator? Gary I thought this was about the air released, not the air that stays in the reservoir. When this thread started some maintained that double the volume at half the pressure could do the same work. I think a lot of the rest of the argument resulted from the informal nature of the discussion. With a blackboard, some sketches and math I think there would be more agreement than disagreement. |
#330
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air regulator (was SCFM ..)
In article ,
Richard J Kinch wrote: Norman Yarvin writes: So suppose you have two containers, one filled with gas, and another empty (filled with vacuum). You connect them, and open the valve between them. The gas molecules bounce around into the newly connected container, so now you have two containers each with half the gas. But the gas temperature is the same (or very close). No, adiabatic expansion, temperature falls. That's what it'd be if you expanded the gas against a piston. Letting the gas pass through a valve is different. The energy that would go into pushing the piston instead goes into reheating the gas. The net result is that the temperature of the gas does not change. (Look up "Joule-Thompson effect" if you want to know the precise story -- really, some gases heat a bit and some cool a bit, under such circumstances. But not much; it's a good approximation to say that the gas does not change in temperature.) -- Norman Yarvin http://yarchive.net |
#331
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air regulator (was SCFM ..)
In article ,
jim rozen wrote: In article , Norman Yarvin says... ... Fermi's book on thermodynamics is much better (as befits one of the greatest physicists of the last century): it spends about the same number of pages on a really good explanation of everything up to and including entropy. It's not easy reading, but it isn't difficult for stupid reasons; it is as easy as the subject permits. Small world. We went out on a treck to a Borders bookstore in a nearby town, and they had that small volume on the shelf. It did not however, have any section on ideal gasses. That's because it deals with ideal gases in every section of the book! For instance, in the section on entropy, it first gives the general argument and definition, then applies that definition to ideal gases to get the formula for the entropy of an ideal gas. I probably should have bought it anyway. Being a Dover book, it is cheap enough that one is not likely to regret the purchase. -- Norman Yarvin http://yarchive.net |
#332
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air regulator (was SCFM ..)
In article , Norman Yarvin says...
(Look up "Joule-Thompson effect" if you want to know the precise story -- really, some gases heat a bit and some cool a bit, under such circumstances. But not much; it's a good approximation to say that the gas does not change in temperature.) Hydrogen has an unusual JT coefficient I think. It will actually heat on expansion. Jim ================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ================================================== |
#333
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air regulator (was SCFM ..)
On Sun, 18 Jan 2004 22:18:48 GMT, Norman Yarvin
wrote: In article , jim rozen wrote: In article , Norman Yarvin says... ... Fermi's book on thermodynamics is much better (as befits one of the greatest physicists of the last century): it spends about the same number of pages on a really good explanation of everything up to and including entropy. It's not easy reading, but it isn't difficult for stupid reasons; it is as easy as the subject permits. Small world. We went out on a treck to a Borders bookstore in a nearby town, and they had that small volume on the shelf. It did not however, have any section on ideal gasses. That's because it deals with ideal gases in every section of the book! For instance, in the section on entropy, it first gives the general argument and definition, then applies that definition to ideal gases to get the formula for the entropy of an ideal gas. I probably should have bought it anyway. Being a Dover book, it is cheap enough that one is not likely to regret the purchase. $9.95 according to the website. 157 pages. I'll be visiting Borders next week. Geez, we should organize a bulk purchase. I think there are quite a few of us that could learn a thing or two from that book. |
#334
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air regulator (was SCFM ..)
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#335
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air regulator (was SCFM ..)
Norman Yarvin writes:
So suppose you have two containers, one filled with gas, and another empty (filled with vacuum). You connect them, and open the valve between them. The gas molecules bounce around into the newly connected container, so now you have two containers each with half the gas. But the gas temperature is the same (or very close). No, adiabatic expansion, temperature falls. That's what it'd be if you expanded the gas against a piston. Letting the gas pass through a valve is different. The energy that would go into pushing the piston instead goes into reheating the gas. The net result is that the temperature of the gas does not change. I would expect that the two containers would end at different temperatures, and thus at least one must have changed. Where the heat from the work of expansion ends up depends on several things, including the geometry of the valve and connections. |
#336
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air regulator (was SCFM ..)
In article ,
Richard J Kinch wrote: Norman Yarvin writes: So suppose you have two containers, one filled with gas, and another empty (filled with vacuum). You connect them, and open the valve between them. The gas molecules bounce around into the newly connected container, so now you have two containers each with half the gas. But the gas temperature is the same (or very close). No, adiabatic expansion, temperature falls. That's what it'd be if you expanded the gas against a piston. Letting the gas pass through a valve is different. The energy that would go into pushing the piston instead goes into reheating the gas. The net result is that the temperature of the gas does not change. I would expect that the two containers would end at different temperatures, and thus at least one must have changed. Where the heat from the work of expansion ends up depends on several things, including the geometry of the valve and connections. Yes, it's the average temperature -- the total amount of energy in the gas -- that will be unchanged. Generally the tank providing the gas will cool, and the other one will heat. Now you could convert some of that temperature difference into mechanical work, by running a Carnot engine between the two containers, but you still wouldn't get as much work from it as you would have done by just expanding the gas against a piston. On a microscopic level, what happens with a valve or a regulator is that the walls are motionless, and bounce gas molecules back at the same speed as they came in with. A piston moving away, on the other hand, bounces gas molecules back more slowly than they came in, since each collision is with a retreating surface; this cools the gas. -- Norman Yarvin http://yarchive.net |
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