[Vo]:Imagine a teakettle
I put some water in a teakettle. I put it on the stove. I turn on the burner, on high. After a while the water in the kettle boils. The steam from the boiling water entirely fills the kettle, pushing out *all* the air. The steam is rushing out the little hole (making an awful whistling noise), *and* the kettle itself is *entirely* filled with steam. The pressure in the kettle is at 1 atmosphere, give or take a few millibars. It's all at 100C, give or take a degree except for the bottom surface of the kettle, which is quite a bit hotter where it contacts the burner. I leave it on the stove and go take a nap, ignoring the whistling noise. After a while, all the water boils to steam. The kettle is still filled with water vapor, of course! But now there's no liquid water left. The stove is still on high, and after a while the bottom of the kettle starts to glow a cheery red. The kettle is still filled with water vapor -- "dry steam" -- and the pressure inside is still 1 atmosphere, give or take a few millibars. What temperature do you suppose the steam inside the kettle is at? Could this be -- gasp! -- an example of superheated steam at 1 atmosphere?? Darn right.
Re: [Vo]:Imagine a teakettle
Stephen A. Lawrence wrote: > The kettle is still filled with water vapor -- "dry steam" -- and the > pressure inside is still 1 atmosphere, give or take a few millibars. > > What temperature do you suppose the steam inside the kettle is at? > Could this be -- gasp! -- an example of superheated steam at 1 atmosphere?? > > Darn right. > Nope. It is 100 deg C. This is well established. The only way you can raise the temperature is to pressurize it. It does not matter what the temperature of kettle surface is. - Jed
Re: [Vo]:Imagine a teakettle
On 02/09/2011 09:43 PM, Jed Rothwell wrote: > Stephen A. Lawrence mailto:sa...@pobox.com>> wrote: (I'm going to put back a few lines you snipped, just for context & clarity:) > After a while, all the water boils to steam. > > The kettle is still filled with water vapor, of course! But now there's > no liquid water left. > > The stove is still on high, and after a while the bottom of the kettle > starts to glow a cheery red. > > > The kettle is still filled with water vapor -- "dry steam" -- and the > pressure inside is still 1 atmosphere, give or take a few millibars. > > > > What temperature do you suppose the steam inside the kettle is at? > > > Could this be -- gasp! -- an example of superheated steam at 1 > atmosphere?? > > Darn right. > [Jed replied:] > > Nope. It is 100 deg C. This is well established. The only way you can > raise the temperature is to pressurize it. It does not matter what the > temperature of kettle surface is. Jed, it's a container, with all the walls at several hundred degrees C or higher; the bottom's in contact with the burner and is probably at about 1000 C. There is nothing inside the container except gas: Gaseous water. Yet you are claiming the gas inside the container can't be hotter than 100 C! That's magic -- it violates the laws of thermodynamics bigtime. How can water vapor be so magical? No other gas behaves that way -- no other material behaves that way!
RE: [Vo]:Imagine a teakettle
Although this discussion thread is really a moot point after it was pointed out that there are 5 PLCs which are controlling the power to the resistive heaters, there's one thing I'd like to point out... Stephen said: "Jed, it's a container, with all the walls at several hundred degrees C or higher; the bottom is in contact with the burner and is probably at about 1000 C." You're assuming that the 'burner' is on full... what if I were to turn the burner to 'simmer'! Then the 'excess' heat would only be slightly higher than what it would take to boil the water... The reactor is very likely engineered so only a small portion of the Ni powder is exposed to whatever conditions are needed to sustain a modest reaction... so using the teakettle/burner jargon, the reactor's 'burner' is on simmer. -Mark
Re: [Vo]:Imagine a teakettle
pV = nRT. If the temperature increases, there must be a corresponding increase in the pressure or the volume (or both). In this tea kettle case, the volume of the steam increases right out the top of the kettle. But the temperature can increase above 100. Sent from my iPhone. On Feb 9, 2011, at 21:43, Jed Rothwell wrote: > Stephen A. Lawrence wrote: > > The kettle is still filled with water vapor -- "dry steam" -- and the > pressure inside is still 1 atmosphere, give or take a few millibars. > > What temperature do you suppose the steam inside the kettle is at? > > Could this be -- gasp! -- an example of superheated steam at 1 atmosphere?? > > Darn right. > > Nope. It is 100 deg C. This is well established. The only way you can raise > the temperature is to pressurize it. It does not matter what the temperature > of kettle surface is. > > - Jed >
Re: [Vo]:Imagine a teakettle
Stephen A. Lawrence wrote: > Jed, it's a container, with all the walls at several hundred degrees C or > higher; the bottom's in contact with the burner and is probably at about > 1000 C. > > There is nothing inside the container except gas: Gaseous water. > > Yet you are claiming the gas inside the container can't be hotter than 100 > C! > If it gets any hotter, the vapor expands and more of it leaves the container. Water vapor at 1 atm cannot be hotter than 100 deg C. The inside of the pot is much hotter of course, since there is heat radiating from the bottom. I meant that if you isolate the steam and take its temperature, it will be 100 deg C. It is, of course, impossible to do that inside the kettle itself. You can only measure some average temperature of the metal and vapor. You have to spray some water in, have it exit the by a hose, and measure the temperature some distance from the hot metal. The kettle temperature might be 500 deg C, but the steam will never be more than 100 deg C, unless you confine it and raise the pressure. > That's magic -- it violates the laws of thermodynamics bigtime. > > How can water vapor be so magical? No other gas behaves that way -- no > other material behaves that way! > All gas behaves that way. There is no violation of thermodynamics, because the vapor occupies more and more space, until eventually it is close to a vacuum inside the kettle. Confine it so that it cannot occupy more space and then the temperature rises. Reduce the pressure around the water to the vacuum of interplanetary space and the liquid boils at any temperature. - Jed
Re: [Vo]:Imagine a teakettle
Hi Jed, What you wrote is true when there is liquid water and steam together in a container - the combination cannot be heated to a temperature higher than 100 deg C without raising the pressure. However once all the liquid has turned to gas there is no longer any limit to what temperature it can be raised to until the molecules dissociate into their component elements. This is analogous to the case of ice and water together in a container. You cannot raise the temperature of the water above 0 deg C until all the ice has melted. Once the ice is melted, then the temperature of the water can easily be raised - until it boils for instance. The reason for both of these is the same - the latent heat of the phase change. In the first case it is the latent heat of vaporization, and in the second it is the latent heat of fusion. While there is a mixture of water in two different phases, then all the energy is absorbed by the water changing phase until it is complete, after that the energy added goes directly into temperature rise instead of driving the phase change. Much the same thing happens as you extract energy to lower the temperature. On 2/10/2011 10:43 AM, Jed Rothwell wrote: Nope. It is 100 deg C. This is well established. The only way you can raise the temperature is to pressurize it. It does not matter what the temperature of kettle surface is.
Re: [Vo]:Imagine a teakettle
wrote: > What you wrote is true when there is liquid water and steam together in a > container - the combination cannot be heated to a temperature higher than > 100 deg C without raising the pressure. However once all the liquid has > turned to gas there is no longer any limit to what temperature it can be > raised to until the molecules dissociate into their component elements. > What would keep the molecules in the kettle, assuming the top or spout is open? What would prevent the gas density from declining indefinitely until it is close to a vacuum? My understanding is that the temperature does not rise as long as the volume is free to expand. In any case, with the real world case of the Rossi device or an actual kettle, you can make the metal as hot as you like, but as long as you keep pumping water in, unconfined steam coming out will be 100 deg C. It may not remove all of the heat, in which case the Rossi device (or kettle) will get hotter and hotter, losing heat by other paths. There is no feeback from the Rossi gadget to the cooling water flow, and no change in the flow rate. There is some other feedback to the electric heater control. I do not know how that works. Anyway, this explains why the steam temperature does not change, even though the Rossi device internal temperature is fluctuating, as you see in one of the graphs. People here have assumed that the cooling water has to be removing all of the heat, keeping everything in balance. There is no indication it is in balance. The unchanging steam temperature does not indicate that the machine is magically supplying just enough heat to keep the steam just above 100 deg C. - Jed
Re: [Vo]:Imagine a teakettle
On 02/10/2011 08:28 AM, Jed Rothwell wrote: > mailto:jwin...@cyllene.uwa.edu.au>> wrote: > > > What you wrote is true when there is liquid water and steam > together in a container - the combination cannot be heated to a > temperature higher than 100 deg C without raising the pressure. > However once all the liquid has turned to gas there is no longer > any limit to what temperature it can be raised to until the > molecules dissociate into their component elements. > > > What would keep the molecules in the kettle, assuming the top or spout > is open? What would prevent the gas density from declining > indefinitely until it is close to a vacuum? > > My understanding is that the temperature does not rise as long as the > volume is free to expand. Your understanding appears to be incomplete. Ideal gases, and, to a good approximation, all real gases, obey the law: PV = nRT That means, for a given quantity of gas, the pressure on the gas, times the volume it occupies, is equal to the number of moles of gas (the quantity), times a constant 'R', times the absolute temperature (degress Kelvin). A "mole" of gas is about 6.02 * 10^23 molecules. If we have a ten gallon tank filled with air, with a small opening, then, when the tank is at room temperature (300 degrees, roughly), we'll have 15 psi * 10 gallons = n0 * R * 300 where the number of *molecules* of gas in the tank is n0 * 6.02 * 10^23. (I leave it to someone else to calculate what the numeric value of "n0" is, and I don't know the value of the constant "R" when we're using pounds per square inch and gallons, so I'll just continue to call it 'R'.) If I then heat the tank to 400 degrees, the air inside will also be heated, by conduction from the walls of the tank. The air will, of course, expand, and we'll probably notice a whistling sound as air rushes out of the tank through the small opening. When it all comes to equilibrium there will be less gas in the tank -- a smaller number of moles. Since we've heated the gas to 4/3 the original temperature, the volume will have expanded by 4/3, so a lot of the gas will have escaped, and the number of moles which will still fit inside the tank will be 3/4 what it was before. So, for the gas which remains inside the tank, we'll have 15 psi * 10 gallons = (3/4)*n0 * R * 400 The temperature has increased by 4/3, the number of moles remaining in the tank has dropped by a factor of 3/4, the pressure and volume are the same as they were to start with, and the equation balances. But you can't make it balance if you assume the temperature didn't increase! As the tank is heated, gas expands and escapes -- you can hear it whistling out. That means "n", the number of moles of gas inside the tank, gets smaller. Since the pressure and the volume inside the tank haven't changed, in order for PV = nRT to continue to balance, the temperature *must* have increased.
Re: [Vo]:Imagine a teakettle
On 2/10/2011 9:28 PM, Jed Rothwell wrote: mailto:jwin...@cyllene.uwa.edu.au>> wrote: What you wrote is true when there is liquid water and steam together in a container - the combination cannot be heated to a temperature higher than 100 deg C without raising the pressure. However once all the liquid has turned to gas there is no longer any limit to what temperature it can be raised to until the molecules dissociate into their component elements. What would keep the molecules in the kettle, assuming the top or spout is open? Well the pressure of the atmosphere into which the gas is exiting will prevent the gas pressure from declining below one atmosphere - allowing the gas to attain any temperature you chose to heat it to and becoming less and less dense until it is indeed close to that of a vacuum. What would prevent the gas density from declining indefinitely until it is close to a vacuum? Here we are speaking of two different quantities that relate to a vacuum - pressure and density. Hot air or helium in a balloon is much less *dense* than the atmosphere, but is never the less at the same *pressure* as the atmosphere. Likewise one could imagine *very* hot air or steam being almost zero *density* (ie close to the density of a vacuum), but still at the same *pressure* as the atmosphere (ie much higher pressure than a vacuum). My understanding is that the temperature does not rise as long as the volume is free to expand. If you allow a gas to expand indefinitely into a vacuum pressure then you are right, it would cool down again until it condensed. But this does require a mechanism to extract its energy as it expands - such as getting it to push on a piston and slowly moving the piston out. If you simply let the gas leak through some sort of throttling valve (like a long thin kettle spout) into a low pressure or vacuum, then as I recall, it usually takes a step down in temperature, but once on the low pressure or vacuum side there is nothing to reduce its temperature further. In fact if my rusty memory serves, some gases under some conditions can exit a throttling valve hotter (in terms of temperature - ie the individual molecules have more kinetic energy) than they were when on the compressed side - this being because the work done by the gas traveling through the expander ends up as frictional heat in the throttling material which gets hot and makes the gas hot also. Sorry my thermodynamics knowledge is so poor - it always seemed a bit too steam-aged to fire my imagination and get me to learn it!
