AGWists are sore losers
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From: Whata Fool on 5 Sep 2007 19:27 On Wed, 05 Sep 2007 15:25:06 -0700, "Phil." <felton(a)princeton.edu> wrote: >Most of Arctic sea ice is formed from freezing sea water rather than >precip I believe. May be? But is there any place near the North Pole that does not get down to freezing at any time of year? Water can melt ice faster than air at the same temperature, and there are probably a lot of other factors involved. For instance, a couple of inches of ice on calm water with a few inches of snow might actually prevent very cold air from freezing more ice. A house feels much warmer on a real cold night if there is new snow on the roof, which apparently slows the loss of heat. It is things like this that make even the well thought out weather or climate news articles questionable at best. Incidently, when I mentioned a dam in the Bering Strait to stop water flow across the Arctic, it is because I don't think a dam any place else would be possible, and a dam any place would stop the flow. But it would also affect weather/climate in England, western Europe, Iceland and Greenland. The contradictory stories about Polar Bears, seals and other mammals would seem to be enough to cause a reasonable person to want more and better information before making major financial decisions. But there are very few people that are able to make decisions that will have much of an effect, and perhaps none of them read this newsgroup.
From: Bill Carter on 5 Sep 2007 21:41 kdthrge(a)yahoo.com wrote: > Instead of sitting around and whining that other people should modify > their modern existence. Even if CO2 caused warming, you have no > program of reducing US emissions in any way which will affect > concentration increases in the air. This fact can be roundly proved. > The simple fact that China and India's growth will completely > overshadow any reductions that we could possibly make is only one. Woo, to think that anyone should modify their 'modern' existence in any way! Other people are stupid enough to pollute, why shouldn't we? The current rate of exploitation of carbon fuels is not sustainable for long, your existence is going to be modified whether you like it or not. Better plan for the future. > So your attempt to promote the passage of laws to stop the ice from > melting is an attempt to commit fraud. In the meantime, enjoy the > euphoria of your delusions and obsession with your HOLY WAR against > society which has no valid basis in climatology. Way too much emotion. If your arguments were persuasive you wouldn't need to behave like this.
From: Bill Ward on 6 Sep 2007 00:14 On Wed, 05 Sep 2007 15:31:44 -0700, kdthrge wrote: > On Sep 5, 12:56 am, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote: >> On Tue, 04 Sep 2007 20:17:04 -0700, kdthrge wrote: >> > On Sep 4, 7:08 pm, Bill Ward . >> >> >> From the kinetic energy of the molecules of the air. >> >> >> > This proves the gases are also transmitting energy according to the >> >> > density of the radiation field of Boltzman Stefan for temperature. >> >> >> There's where you lose me. It seems to me the gas is equilibrating >> >> with the plate, then the plate radiates the energy. The radiation is >> >> emitted from the plate, not the gas. >> >> > But it must transfer through the gas in order to radiate from the >> > metal. >> >> > It is impossible that the kinetic energy of the gases is being >> > transmitted into the metal, >> >> Why do you say that? The molecules of gas are pounding on the molecules >> of metal, and transferring KE. What about that is "impossible"? >> >> > and that on the other side of the metal there is radiation of Boltzman >> > Stefan from the transference of energy in this fashion. This proves >> > the existence of the energy of the radiation field. >> >> There's radiation from the metal, but not from the gas. The gas to >> metal transfer is by direct molecular impacts. >> >> >> >> > The average kinetic energy of the molecules of a gas is kT. At 300K >> > this is about .025eV. For 1 mole this is 2490 Joules. This is the >> > kinetic energy of all the velocities of every molecule in 1 mole. This >> > is the sum total of all of the energy of all of the velocities of the >> > mole of gas. >> >> I don't think those numbers are exactly correct, (IIRC, KE is 3/2 kT) >> but I'd rather not tackle that now and stay on a qualitative level. >> >> >> >> > At 300K, 459 Joules per second per sq meter is leaving the surface of >> > the metal. This means in 5.4 seconds, all of the energy of all of the >> > velocities of the molecules in 1 mole of gas must be delivered through >> > the metal of 1 sq meter from the gas for the metal to maintain this >> > temperature. >> >> Don't forget that the T will drop rapidly unless energy in coming in >> from somewhere else. And you're assuming a 0K black body sink. >> >> > This mole of gas if 1 meter square would be .38 meter deep. >> >> > If one is using a monatomic gas at constant volume, the heat capacity >> > of this gas is 3/2RT or 3735 Joules. Although not exact, this is the >> > total heat in one mole of monatomic gas above absolute zero. >> >> > At 460 Joules per second per sq meter, this energy would be lost in >> > 8.1 seconds. >> >> Only if you were radiating into 0K, and no energy was coming in from the >> source. >> >> > What if you cut the source of heat into the gas and evaluated the time >> > of total heat loss, rate of radiated energy loss with decreasing >> > temperature and how much time this takes? >> >> > You see nothing wrong with these numbers for the kinetic energy of the >> > velocities and the quantity of energy that is radiated, or that it is >> > possible for the gas to deliver this energy at this rate to the metal >> > by collisional energy? >> >> Yes, actually 460 Wm-2 seems plausible if the gas is hot enough and the >> metal thin enough. Remember we're off equilibrium, because heat is >> flowing through the system. The amount of heat flowing is proportional >> to the temperature drops. >> >> >> >> > The idea of the metal and the transference of heat to and through the >> > metal is to prove the idea false that gases do not obey to Boltzman >> > Stefan for temperature and energy. >> > The existence of the radiation field is proved. There is no other >> > possible means for this energy to be transferred to the metal. >> >> I don't see the proof. Why can't it be molecular collisions? >> >> >> >> > It is impossible that the energy of the velocities of the molecules is >> > transfering the energy to the metal which then leaves the other side >> > at the rate that it does. >> >> > Let me see some numbers in Joules that show this to be possible. Or >> > show me where this figuring is wrong and where the energy comes from >> > that is radiating from the metal if it is at 300K. The room could be >> > cooled, and the temperature in the duct adjusted so that the exterior >> > of the metal was at 300K. This is to reduce the radiation from the >> > room on the outside of the metal for the accounting of the energy that >> > is being radiated. >> >> If the room (sink) were at 300 K you'd be at equilibrium, with no >> radiative transport. >> >> >> >> > At this point I damn sure need to see your accounting for the energy >> > supposedly transfered into the metal by the collisions of the >> > molecules. >> >> It comes from the gas, which presumably gets it from a similar plate on >> the other end of your cell, which gets it from ... >> >> > Or maybe you could explain your viewpoint and education that can >> > quantify this energy according to collisional transfer of energy. >> >> I'm not really qualified. What I'd recommend is that you take a look at >> what hyperphysics has to say about kinetic theory: >> >> http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/ktcon.html#c1 >> >> I have to warn you that they have unfortunately fallen for the AGW scam, >> but most of the traditional physics looks OK to me. It's easy to follow >> the hyperlinks to browse around to related areas. >> >> >> > If the pressure is decreased radically, there becomes too few >> >> > molecules to adequately absorb radiation and re-emit this >> >> > radiation. >> >> >> If you reduce the gas pressure, you reduce the temperature, and must >> >> add energy to come back to the same temperature. You haven't yet >> >> shown that there's any radiation interacting with the gas. >> >> > Of course you do. But now at the same temperature, the transfer of >> > energy is very nearly the same, although the number of collisions is >> > reduced. No viable evidence that the collisions are the sole means >> > that energy is transfering to the metal. >> >> But many experiments show the conductivity is actually proportional to >> the pressure. Thats how Pirani vacuum gages work.- Hide quoted text - >> >> - Show quoted text - > > If you see that the kinetic motions of the molecules can transfer the > energy to the metal so that the metal radiates as it does, let's look at > higher temperatures. > > At 1000K, RT = 8315 Joules > 3/2 RT = 12472 Joules > Stefan Boltzman = 56,700 Joules per second per sq meter The frequency of > highest intenstity of the radiation is 2.892um,, or .428 eV > kT is equal to .0864eV or a photon of 14.345um > > So the mean kinetic energy if considered as 3/2 kT is .1296 eV > > If the gas within the duct is maintained at 1000K, it will bring the > temperature of the metal plate to this temperature at which time it will > radiate at these specifics. At higher energies you need to look at the appropriate part of the absorption spectrum. At higher temperatures gases begin to radiate. But if you watch the ignition sequence on the shuttle main engines, you'll see the exhaust go transparent after combustion stabilizes. You can look through the hot, dense gas and clearly see the opposite side of the bell, which is white hot. The flame is essentially invisible, so it's not radiating much. > This rate of radiation is enough to radiate the energy of 1 mole of > monatomic gas, (3/2 RT) in .21 seconds. > > So you are saying that the kinetic energy of the molecules which has an > average equal to a photon of 14um is transfering energy to the metal to > produce the photons of greatest intensity at 2.898 um. And that this > quantity of gas with this average kinetic energy of the molecules is > capable of producing the 56,700 Joules per second per meter? I'm not necessarily saying the numbers are correct, because it would take too long for me to check them. I think Hyperphysics has a radiative calculator that speeds up playing around with numerical examples like that > > Should we go to higher temperatures? > O2 and N2 can easily be heated to higher temperatures in which they > radiate in the visible. So now we place the gas in a metal container > with a glass window. > > Only rarified gases radiate in the specific intervals of spectra. > > All gases under pressure emit the same continous spectra through the > visible frequencies. So at a little higher temperature, intensity > maximum will be clearly in the visible, while average kinetic energy of > the molecules is still in the infrared. > > This means that when a molecule collides when at average velocity, if it > loses ALL of it's energy into a photon, this will still be 4.95 times > less than the energy of highest intensity photons. > > How do the oxygen and nitrogen gas molecules attain the energy in which > they are clearly radiating in the continous spectra in the visible, if > they are not absorbing this energy in the infrared? > > The light leaving through the window is a quantity of energy. This loss > of energy to the system, the loss of the heat through the rest of the > container cannot be from the velocities and collisions of the molecules. > > The rate of replacement of this energy from the heat source through the > gas to the molecules that are losing their energy to the environment > cannot be quantified without the concept of the radiation field. The > energy that is transfered through the gas to the metal is of this > radiation field and is radiated as packets of energy from the electron > membranes or oscillator of the molecules. You might want to browse through hyperphysics, as they will be able to help you a lot more than I can. It's getting way over my head.
From: kdthrge on 6 Sep 2007 05:53 On Sep 5, 11:14 pm, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote: > On Wed, 05 Sep 2007 15:31:44 -0700, kdthrge wrote: > > On Sep 5, 12:56 am, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote: > >> On Tue, 04 Sep 2007 20:17:04 -0700, kdthrge wrote: > >> > On Sep 4, 7:08 pm, Bill Ward . > > >> >> From the kinetic energy of the molecules of the air. > > >> >> > This proves the gases are also transmitting energy according to the > >> >> > density of the radiation field of Boltzman Stefan for temperature. > > >> >> There's where you lose me. It seems to me the gas is equilibrating > >> >> with the plate, then the plate radiates the energy. The radiation is > >> >> emitted from the plate, not the gas. > > >> > But it must transfer through the gas in order to radiate from the > >> > metal. > > >> > It is impossible that the kinetic energy of the gases is being > >> > transmitted into the metal, > > >> Why do you say that? The molecules of gas are pounding on the molecules > >> of metal, and transferring KE. What about that is "impossible"? > > >> > and that on the other side of the metal there is radiation of Boltzman > >> > Stefan from the transference of energy in this fashion. This proves > >> > the existence of the energy of the radiation field. > > >> There's radiation from the metal, but not from the gas. The gas to > >> metal transfer is by direct molecular impacts. > > >> > The average kinetic energy of the molecules of a gas is kT. At 300K > >> > this is about .025eV. For 1 mole this is 2490 Joules. This is the > >> > kinetic energy of all the velocities of every molecule in 1 mole. This > >> > is the sum total of all of the energy of all of the velocities of the > >> > mole of gas. > > >> I don't think those numbers are exactly correct, (IIRC, KE is 3/2 kT) > >> but I'd rather not tackle that now and stay on a qualitative level. > > >> > At 300K, 459 Joules per second per sq meter is leaving the surface of > >> > the metal. This means in 5.4 seconds, all of the energy of all of the > >> > velocities of the molecules in 1 mole of gas must be delivered through > >> > the metal of 1 sq meter from the gas for the metal to maintain this > >> > temperature. > > >> Don't forget that the T will drop rapidly unless energy in coming in > >> from somewhere else. And you're assuming a 0K black body sink. > > >> > This mole of gas if 1 meter square would be .38 meter deep. > > >> > If one is using a monatomic gas at constant volume, the heat capacity > >> > of this gas is 3/2RT or 3735 Joules. Although not exact, this is the > >> > total heat in one mole of monatomic gas above absolute zero. > > >> > At 460 Joules per second per sq meter, this energy would be lost in > >> > 8.1 seconds. > > >> Only if you were radiating into 0K, and no energy was coming in from the > >> source. > > >> > What if you cut the source of heat into the gas and evaluated the time > >> > of total heat loss, rate of radiated energy loss with decreasing > >> > temperature and how much time this takes? > > >> > You see nothing wrong with these numbers for the kinetic energy of the > >> > velocities and the quantity of energy that is radiated, or that it is > >> > possible for the gas to deliver this energy at this rate to the metal > >> > by collisional energy? > > >> Yes, actually 460 Wm-2 seems plausible if the gas is hot enough and the > >> metal thin enough. Remember we're off equilibrium, because heat is > >> flowing through the system. The amount of heat flowing is proportional > >> to the temperature drops. > > >> > The idea of the metal and the transference of heat to and through the > >> > metal is to prove the idea false that gases do not obey to Boltzman > >> > Stefan for temperature and energy. > >> > The existence of the radiation field is proved. There is no other > >> > possible means for this energy to be transferred to the metal. > > >> I don't see the proof. Why can't it be molecular collisions? > > >> > It is impossible that the energy of the velocities of the molecules is > >> > transfering the energy to the metal which then leaves the other side > >> > at the rate that it does. > > >> > Let me see some numbers in Joules that show this to be possible. Or > >> > show me where this figuring is wrong and where the energy comes from > >> > that is radiating from the metal if it is at 300K. The room could be > >> > cooled, and the temperature in the duct adjusted so that the exterior > >> > of the metal was at 300K. This is to reduce the radiation from the > >> > room on the outside of the metal for the accounting of the energy that > >> > is being radiated. > > >> If the room (sink) were at 300 K you'd be at equilibrium, with no > >> radiative transport. > > >> > At this point I damn sure need to see your accounting for the energy > >> > supposedly transfered into the metal by the collisions of the > >> > molecules. > > >> It comes from the gas, which presumably gets it from a similar plate on > >> the other end of your cell, which gets it from ... > > >> > Or maybe you could explain your viewpoint and education that can > >> > quantify this energy according to collisional transfer of energy. > > >> I'm not really qualified. What I'd recommend is that you take a look at > >> what hyperphysics has to say about kinetic theory: > > >>http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/ktcon.html#c1 > > >> I have to warn you that they have unfortunately fallen for the AGW scam, > >> but most of the traditional physics looks OK to me. It's easy to follow > >> the hyperlinks to browse around to related areas. > > >> >> > If the pressure is decreased radically, there becomes too few > >> >> > molecules to adequately absorb radiation and re-emit this > >> >> > radiation. > > >> >> If you reduce the gas pressure, you reduce the temperature, and must > >> >> add energy to come back to the same temperature. You haven't yet > >> >> shown that there's any radiation interacting with the gas. > > >> > Of course you do. But now at the same temperature, the transfer of > >> > energy is very nearly the same, although the number of collisions is > >> > reduced. No viable evidence that the collisions are the sole means > >> > that energy is transfering to the metal. > > >> But many experiments show the conductivity is actually proportional to > >> the pressure. Thats how Pirani vacuum gages work.- Hide quoted text - > > >> - Show quoted text - > > > If you see that the kinetic motions of the molecules can transfer the > > energy to the metal so that the metal radiates as it does, let's look at > > higher temperatures. > > > At 1000K, RT = 8315 Joules > > 3/2 RT = 12472 Joules > > Stefan Boltzman = 56,700 Joules per second per sq meter The frequency of > > highest intenstity of the radiation is 2.892um,, or .428 eV > > kT is equal to .0864eV or a photon of 14.345um > > > So the mean kinetic energy if considered as 3/2 kT is .1296 eV > > > If the gas within the duct is maintained at 1000K, it will bring the > > temperature of the metal plate to this temperature at which time it will > > radiate at these specifics. > > At higher energies you need to look at the appropriate part of the > absorption spectrum. At higher temperatures gases begin to radiate. > But if you watch the ignition sequence on the shuttle main engines, you'll > see the exhaust go transparent after combustion stabilizes. You can look > through the hot, dense gas and clearly see the opposite side of the bell, > which is white hot. The flame is essentially invisible, so it's not > radiating much. > > > This rate of radiation is enough to radiate the energy of 1 mole of > > monatomic gas, (3/2 RT) in .21 seconds. > > > So you are saying that the kinetic energy of the molecules which has an > > average equal to a photon of 14um is transfering energy to the metal to > > produce the photons of greatest intensity at 2.898 um. And that this > > quantity of gas with this average kinetic energy of the molecules is > > capable of producing the 56,700 Joules per second per meter? > > I'm not necessarily saying the numbers are correct, because it would take > too long for me to check them. I think Hyperphysics has a radiative > calculator that speeds up playing around with numerical examples like that > > > > > > > > > Should we go to higher temperatures? > > O2 and N2 can easily be heated to higher temperatures in which they > > radiate in the visible. So now we place the gas in a metal container > > with a glass window. > > > Only rarified gases radiate in the specific intervals of spectra. > > > All gases under pressure emit the same continous spectra through the > > visible frequencies. So at a little higher temperature, intensity > > maximum will be clearly in the visible, while average kinetic energy of > > the molecules is still in the infrared. > > > This means that when a molecule collides when at average velocity, if it > > loses ALL of it's energy into a photon, this will still be 4.95 times > > less than the energy of highest intensity photons. > > > How do the oxygen and nitrogen gas molecules attain the energy in which > > they are clearly radiating in the continous spectra in the visible, if > > they are not absorbing this energy in the infrared? > > > The light leaving through the window is a quantity of energy. This loss > > of energy to the system, the loss of the heat through the rest of the > > container cannot be from the velocities and collisions of the molecules. > > > The rate of replacement of this energy from the heat source through the > > gas to the molecules that are losing their energy to the environment > > cannot be quantified without the concept of the radiation field. The > > energy that is transfered through the gas to the metal is of this > > radiation field and is radiated as packets of energy from the electron > > membranes or oscillator of the molecules. > > You might want to browse through hyperphysics, as they will be able to > help you a lot more than I can. It's getting way over my head.- Well let's analyse the basics which clearly prove 'prince' Phil a charlatan and the other theorists of AGW completely fraudulent in their theory and in their claim to have any validity at all in physics or any knowledge whatsoever of gases, temperature, energy or the subject of their concern. Pressure and temperature are directly proportional. Average velocity increases as a square root to the mean kinetic energy due to 1/2mv^2. Average velocity and number of collisons in unit time are directly proportional. Mean kinetic energy increases directly proportional to temperature as kT. So from 300K to 1000K is 700deg temperature increase which is a 233% increase in temperature and pressure. Pressure is a product of average velocity x number of collisions. So,, square root of 1000 - square root of 300 = increase in average velocity and also increase in number of collisions in a specific time interval. 31.6 - 17.3 = 14.3,,,14.3 / 17.3 = .8266 or 83% Average velocity and number of collisons both increase by 83% with temperature increase from 300K to 1000K. However, the energy radiating from the piece of metal in the air duct increases from 460Wm-2 to 56,700Wm-2. 56,700 - 460 = 56,240,, 56,240 / 460 = 122.26 or 12,226% The energy radiated per second from the metal which is recieving it's energy through the gas increases by 12,226%. The average velocity increases by 83% The number of collisions per time increases 83% The energy transfered increases by, 12,226% due to the Stefans Law and the increase of energy as a fourth power to absolute temperature. It is impossible that the energy is being transfered by the collisons and kinetic energies of the molecules. The existence of a radiation field within the gas is entirely proved. The density of the energy of this field is the determination of temperature and energy which is transfered. The gas molecules of O2 and N2 absorb the infrared frequencies in order to transfer the energy. They cannot possibly be tranfering energy by their mechanical motions as is claimed by the fraudulent theory of 'grenhouse gases'. The stated theory of the 'grenhouse gases' which entirely depends on their false idea that O2 and N2 are transparent to infrared frequencies and that these frequencies are absorbed by specific gases is complete fraud. KDeatherage CO2Phobia is a psychological disease. Seek professional help.
From: kdthrge on 6 Sep 2007 05:58
On Sep 5, 5:31 pm, kdth...(a)yahoo.com wrote: > On Sep 5, 12:56 am, Bill Ward <bw...(a)REMOVETHISix.netcom.com> wrote: > > > > > > > On Tue, 04 Sep 2007 20:17:04 -0700, kdthrge wrote: > > > On Sep 4, 7:08 pm, Bill Ward . > > > >> From the kinetic energy of the molecules of the air. > > > >> > This proves the gases are also transmitting energy according to the > > >> > density of the radiation field of Boltzman Stefan for temperature. > > > >> There's where you lose me. It seems to me the gas is equilibrating with > > >> the plate, then the plate radiates the energy. The radiation is emitted > > >> from the plate, not the gas. > > > > But it must transfer through the gas in order to radiate from the metal. > > > > It is impossible that the kinetic energy of the gases is being transmitted > > > into the metal, > > > Why do you say that? The molecules of gas are pounding on the molecules > > of metal, and transferring KE. What about that is "impossible"? > > > > and that on the other side of the metal there is radiation > > > of Boltzman Stefan from the transference of energy in this fashion. This > > > proves the existence of the energy of the radiation field. > > > There's radiation from the metal, but not from the gas. The gas to metal > > transfer is by direct molecular impacts. > > > > The average kinetic energy of the molecules of a gas is kT. At 300K this > > > is about .025eV. For 1 mole this is 2490 Joules. This is the kinetic > > > energy of all the velocities of every molecule in 1 mole. This is the > > > sum total of all of the energy of all of the velocities of the mole of > > > gas. > > > I don't think those numbers are exactly correct, (IIRC, KE is 3/2 kT) > > but I'd rather not tackle that now and stay on a qualitative level. > > > > At 300K, 459 Joules per second per sq meter is leaving the surface of > > > the metal. This means in 5.4 seconds, all of the energy of all of the > > > velocities of the molecules in 1 mole of gas must be delivered through > > > the metal of 1 sq meter from the gas for the metal to maintain this > > > temperature. > > > Don't forget that the T will drop rapidly unless energy in coming in from > > somewhere else. And you're assuming a 0K black body sink. > > > > This mole of gas if 1 meter square would be .38 meter deep. > > > > If one is using a monatomic gas at constant volume, the heat capacity of > > > this gas is 3/2RT or 3735 Joules. Although not exact, this is the total > > > heat in one mole of monatomic gas above absolute zero. > > > > At 460 Joules per second per sq meter, this energy would be lost in 8.1 > > > seconds. > > > Only if you were radiating into 0K, and no energy was coming in from the > > source. > > > > What if you cut the source of heat into the gas and evaluated the time > > > of total heat loss, rate of radiated energy loss with decreasing > > > temperature and how much time this takes? > > > > You see nothing wrong with these numbers for the kinetic energy of the > > > velocities and the quantity of energy that is radiated, or that it is > > > possible for the gas to deliver this energy at this rate to the metal by > > > collisional energy? > > > Yes, actually 460 Wm-2 seems plausible if the gas is hot enough and the > > metal thin enough. Remember we're off equilibrium, because heat is flowing > > through the system. The amount of heat flowing is proportional to the > > temperature drops. > > > > The idea of the metal and the transference of heat to and through the > > > metal is to prove the idea false that gases do not obey to Boltzman > > > Stefan for temperature and energy. > > > The existence of the radiation field is proved. There is no other > > > possible means for this energy to be transferred to the metal. > > > I don't see the proof. Why can't it be molecular collisions? > > > > It is impossible that the energy of the velocities of the molecules is > > > transfering the energy to the metal which then leaves the other side at > > > the rate that it does. > > > > Let me see some numbers in Joules that show this to be possible. Or show > > > me where this figuring is wrong and where the energy comes from that is > > > radiating from the metal if it is at 300K. The room could be cooled, and > > > the temperature in the duct adjusted so that the exterior of the metal > > > was at 300K. This is to reduce the radiation from the room on the > > > outside of the metal for the accounting of the energy that is being > > > radiated. > > > If the room (sink) were at 300 K you'd be at equilibrium, with no > > radiative transport. > > > > At this point I damn sure need to see your accounting for the energy > > > supposedly transfered into the metal by the collisions of the molecules. > > > It comes from the gas, which presumably gets it from a similar plate on > > the other end of your cell, which gets it from ... > > > > Or maybe you could explain your viewpoint and education that can > > > quantify this energy according to collisional transfer of energy. > > > I'm not really qualified. What I'd recommend is that you take a look at > > what hyperphysics has to say about kinetic theory: > > >http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/ktcon.html#c1 > > > I have to warn you that they have unfortunately fallen for the AGW scam, > > but most of the traditional physics looks OK to me. It's easy to follow > > the hyperlinks to browse around to related areas. > > > >> > If the pressure is decreased radically, there becomes too few > > >> > molecules to adequately absorb radiation and re-emit this radiation. > > > >> If you reduce the gas pressure, you reduce the temperature, and must > > >> add energy to come back to the same temperature. You haven't yet shown > > >> that there's any radiation interacting with the gas. > > > > Of course you do. But now at the same temperature, the transfer of > > > energy is very nearly the same, although the number of collisions is > > > reduced. No viable evidence that the collisions are the sole means that > > > energy is transfering to the metal. > > > But many experiments show the conductivity is actually proportional to the > > pressure. Thats how Pirani vacuum gages work.- Hide quoted text - > > > - Show quoted text - > > If you see that the kinetic motions of the molecules can transfer the > energy to the metal so that the metal radiates as it does, let's look > at higher temperatures. > > At 1000K, RT = 8315 Joules > 3/2 RT = 12472 Joules > Stefan Boltzman = 56,700 Joules per second per sq meter > The frequency of highest intenstity of the radiation is > 2.892um,, or .428 eV > kT is equal to .0864eV or a photon of 14.345um > > So the mean kinetic energy if considered as 3/2 kT is .1296 eV > > If the gas within the duct is maintained at 1000K, it will bring the > temperature of the metal plate to this temperature at which time it > will radiate at these specifics. > > This rate of radiation is enough to radiate the energy of 1 mole of > monatomic gas, (3/2 RT) in .21 seconds. > > So you are saying that the kinetic energy of the molecules which has > an average equal to a photon of 14um is transfering energy to the > metal to produce the photons of greatest intensity at 2.898 um. And > that this quantity of gas with this average kinetic energy of the > molecules is capable of producing the 56,700 Joules per second per > meter? > > Should we go to higher temperatures? > O2 and N2 can easily be heated to higher temperatures in which they > radiate in the visible. So now we place the gas in a metal container > with a glass window. > > Only rarified gases radiate in the specific intervals of spectra. > > All gases under pressure emit the same continous spectra through the > visible frequencies. So at a little higher temperature, intensity > maximum will be clearly in the visible, while average kinetic energy > of the molecules is still in the infrared. > > This means that when a molecule collides when at average velocity, if > it loses ALL of it's energy into a photon, this will still be 4.95 > times less than the energy of highest intensity photons. > > How do the oxygen and nitrogen gas molecules attain the energy in > which they are clearly radiating in the continous spectra in the > visible, if they are not absorbing this energy in the infrared? > > The light leaving through the window is a quantity of energy. This > loss of energy to the system, the loss of the heat through the rest of > the container cannot be from the velocities and collisions of the > molecules. > > The rate of replacement of this energy from the heat source through > the gas to the molecules that are losing their energy to the > environment cannot be quantified without the concept of the radiation > field. The energy that is transfered through the gas to the metal is > of this radiation field and is radiated as packets of energy from the > electron membranes or oscillator of the molecules. > Well let's analyse the basics which clearly prove 'prince' Phil a charlatan and the other theorists of AGW completely fraudulent in their theory and in their claim to have any validity at all in physics or any knowledge whatsoever of gases, temperature, energy or the subject of their concern. Pressure and temperature are directly proportional. Average velocity increases as a square root to the mean kinetic energy due to 1/2mv^2. Average velocity and number of collisons in unit time are directly proportional. Mean kinetic energy increases directly proportional to temperature as kT. So from 300K to 1000K is 700deg temperature increase which is a 233% increase in temperature and pressure. Pressure is a product of average velocity x number of collisions. So,, square root of 1000 - square root of 300 = increase in average velocity and also increase in number of collisions in a specific time interval. 31.6 - 17.3 = 14.3,,,14.3 / 17.3 = .8266 or 83% Average velocity and number of collisons both increase by 83% with temperature increase from 300K to 1000K. However, the energy radiating from the piece of metal in the air duct increases from 460Wm-2 to 56,700Wm-2. 56,700 - 460 = 56,240,, 56,240 / 460 = 122.26 or 12,226% The energy radiated per second from the metal which is recieving it's energy through the gas increases by 12,226%. The average velocity increases by 83% The number of collisions per time increases 83% The energy transfered increases by, 12,226% due to the Stefans Law and the increase of energy as a fourth power to absolute temperature. It is impossible that the energy is being transfered by the collisons and kinetic energies of the molecules. The existence of a radiation field within the gas is entirely proved. The density of the energy of this field is the determination of temperature and energy which is transfered. The gas molecules of O2 and N2 absorb the infrared frequencies in order to transfer the energy. They cannot possibly be tranfering energy by their mechanical motions as is claimed by the fraudulent theory of 'grenhouse gases'. The stated theory of the 'grenhouse gases' which entirely depends on their false idea that O2 and N2 are transparent to infrared frequencies and that these frequencies are absorbed by specific gases is complete fraud. KDeatherage CO2Phobia is a psychological disease. Seek professional help. |