Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: paul.fr on 11/05/2007 21:54:28
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On a sunny day, i can burn paper with my magnifying glass. If i had a really good glass, could i burn through steel? or maybe the earths core!!!
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You can certainly burn through steel (though, it is not just how sunny it is, but how big the magnifying glass).
In theory you could create enough concentration of heat that you could melt the surface of the Earth, but since you are only focusing one one point, you will only melt the absolute surface, and not the layer beneath (at least, not without moving the magnifying glass to shift the point of focus of the light, and continue moving is forward as you melt each level of the surface layer). You would ofcourse need a very very large magnifying glass to collect enough light to do this with.
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Excellent. How big would it need to be, to burn a hole throuh steel?
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I'm not sure about burning a hole.. But to melt it.. I assume you'll only need about the size of ... 1m squared?
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On a sunny day, i can burn paper with my magnifying glass. If i had a really good glass, could i burn through steel? or maybe the earths core!!!
The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.
So it's not so easy to reach the steel melting point.
However, you could use a mirror, instead; I assume it would be better for that purpose.
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The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.
I agree that the aberrations will cause a defocusing of the lens, but I cannot see that it is not dependent on lens diameter, since the larger the lens, the larger amount of light must be being focused (in effect, you have a wider aperture lens). Also, the larger the lens, the less are likely to be the aberrations, since most of the aberrations tend to be towards the outer parts of the lens, and a larger lens will also have a larger core region (OK, you have mentioned that you would get fewer spherical aberrations, but I would have thought that also you would have less chromatic aberrations - and I am not sure why you think a larger lens would have a longer focal length - that will depend on the curvature, but not the diameter).
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You can correct for spherical and chromatic aberation to a fairly good degree. On the other hand the size of the focussed spot depends on the size of the sun and the focal length of the lens. A shorter focal length will give a smaller spot but it's difficult to make big lenses with short focal lengths (there are limits to how big a lens you can make for a given focal length) and the bigger the lens the more difficult it is to correct for spherical aberation.
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It's just got to depend upon the actual size of the steel object and the total amount of energy being focussed on 'the spot', more than how small you can make the spot.
"A good big un will beat a good little un" any day.
The steel will conduct away so much heat and reduce the final temperature of the hot spot so, unless you can produce a large power flow into your target area, you have no chance of melting a large object.
In the end, you can only rely on less than 1kW per m squared, whatever you want to do.
Electric arc welders use a few kW of power, in practice.
A huge lens - 1m squared - will direct the best part of 1kW on your spot on a sunny day and a bit of aberration won't matter much at all. 1kW would melt a small hole in a thin sheet of steel easily.
It's a different matter if you just want to melt a tiny sample, particularly if you could keep it in an evacuated glass jar (optically flat face, of course). In that case you could melt a small piece of steel with less than 1W - so a lens of area 0.001m squared would do - just a few cm across. The basis for this argument is that a small light bulb filament (1W power) can't be made of steel, but has to be made of tungsten to avoid it melting.
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The light intensity you can focus on a small area doesn't depend on lens diameter, and is limited from lens aberrations, first chromatic and then spherical aberration; If you take a large lens with high focal distance, aberrations are less, but the effect of chromatic aberration, for that purpose, is actually increased because of the higher distance from the various wavelenghts' focal planes.
I agree that the aberrations will cause a defocusing of the lens, but I cannot see that it is not dependent on lens diameter, since the larger the lens, the larger amount of light must be being focused (in effect, you have a wider aperture lens). Also, the larger the lens, the less are likely to be the aberrations, since most of the aberrations tend to be towards the outer parts of the lens, and a larger lens will also have a larger core region (OK, you have mentioned that you would get fewer spherical aberrations, but I would have thought that also you would have less chromatic aberrations - and I am not sure why you think a larger lens would have a longer focal length - that will depend on the curvature, but not the diameter).
If you have a bigger lens, you will have a greater amount of heat transferred to the object, so the object won't cool as fast, giving away its heat, we agree on this.
However the light's intensity incident on the object's surface, is the same, if the lens optical quality is the same, because you cannot focus light onto a disk with a diameter less than k*D/f, where D is the lens diameter, f the focal lenght and k a constant, because of aberrations; the theoretical limit is due to the sun's disk, as Bored chemist said. So the rate (lens area)/(focused disk area) = incident light's intensity, is the same. Making an analogy, it's like heating an object with a small or a big bunsen with the same flame's temperature: you cannot reach a much higher temperature of the object, just because the flame's temperature is the same.
To increase the rate (lens area)/(focused disk area) you have to change material (higher refractive index) or decrease the focal lenght increasing the lens curvature (higher aberrations) or increase the optical quality.
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Yes - that is reasonable but, practically, the spot size is not the only relevant factor. As soon as the image is focussed, it starts to heat up the material and there is a temperature gradient as around the spot as heat is transferred outward. The actual temperature of the spot, in terms of the spectrum of the radiation arriving at the spot is that of the image of the Sun and that's quite hot enough to melt most things! That is the limiting temperature of any object you want to heat up. Isn't it all about Stefan's law?
What limits the actual temperature reached is the rate of transfer of heat away from it.
This is turning into a typical Naked Sci type discussion - we keep shifting the goal posts.
But, In practical terms, it's just got to work best with a lot of energy from a big lens.
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Hi people, bit of fun... Somebody mentioned using a mirror instead. So the thought occurred to me earlier, just for fun, when reading about the National Ignition Facility project, that maybe the 4.2 billion dollars spent on this could be better used simply by finding a big mountain somewhere, just for fun, and building a humungous 2-kilometer wide parabolic mirror into the side of it and having a moveable target suspended by a crane, following the sun, and generating fusion on the target. Just for fun, I wondered what the upper limit of temperature was that could be reached, in theory, with such a device ? Anyone know how many joules per meter squared the sun can normally deliver to the ground on a nice sunny day? No one has offered any figures, so let's take a guess. My 10 cm-diameter magnifying glass can easily burn paper with a dot that is, at a guess, about 100 times smaller than the glass itself. If the temperature generated is say 250 degrees, then we can start having some fun. Let's assume for the sake of argument that a 10cm parabolic mirror would generate the same amount of heat; I suspect it would be more, but let's assume this to make things easier. My glass / mirror is 5 squared times PI cm squared, which is 78.5 . So an area of 78.5 cm squared gives you a manageable dot of light at 250 degrees celcius. So, a mirror of two kilometers wide = 2000 meters = 200000 centimeters. Square this and times by PI, and you get and area of 125663704000 cm squared. Now, divide this by the area of my magnifying glass, 78.5 cm squared, and you learn that the big mirror is 1600811515.9 times bigger in area. Times this by 250 degrees and you get a temperature of 400,202,878,980 or about 400 thousand million degrees, easily enough to start fusion! Now, obviously, our great scientists would have tried this, I guess, unless they spotted what would be to them an obvious flaw in the scheme. Anybody care to put me right? :-) thanks :-)
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The temperature doesn't scale with the area of the glass.
The limiting temperature happens when the hot thing is losing energy by radiation (etc) as fast as it is gaining it from the light. The radiation rises with the 4th power of the temperature and with the area of the object. The power varies as the area of the magnifying glass. If the object being heated is big then the temperature rises as the square root of the size of the lens (I think) but if you want maximum temperature the object heated must be just bigger than the sizee of the image of the sun. That is roughly proportional to the size of the lens so overall a huge lens doesn't get much benefit in terms of maximum temperature. it does make a difference to the power available
Here's an article about a pretty big mirror (easier to work with than a big lens)
http://www.time.com/time/magazine/article/0,9171,909204,00.html
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I would have though the maximum temperature would be about 6000K, which is the colour temperature of sunlight.
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Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?
For example, it's difficult to ascribe a 'colour temperature' to a near-monochromatic light source (e.g. a LASER) but it is certainly possible to use a focussed monochromatic beam to heat a target.
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Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?
For example, it's difficult to ascribe a 'colour temperature' to a near-monochromatic light source (e.g. a LASER) but it is certainly possible to use a focussed monochromatic beam to heat a target.
Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?
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Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?
Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.
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It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature
Agreed.
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"Surely this would have more to do with the absorbtion/radiation characteristics of the target than of the source of light used?"
All things at all wavelengthts are exactly as good at radiating as they are at absorbing so the effet evens out.
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All things at all wavelengthts are exactly as good at radiating as they are at absorbing so the effect evens out.
A bit of an odd thing to say: What you are describing is an ideal 'black body'. Real objects don't achieve this.
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Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.
This is certainly true, but I am not sure how it relates to the argument.
We say that the surface of the Sun is approx 6000K, because that is the temperature of the light emanating from the surface. We know that the temperature beneath the surface can reach millions of K, but that is not the radiation we receive, so we know that the temperature on the surface is not as hot an the depths of the Sun are.
Yes, you can put a filter in-front of an object, and that will cause the system as a whole to cool on its exterior (beyond that filter), but it does not prevent the interior of the system from being hotter.
The question is, should not the radiation emitted from a 6000K body should be able to induce a temperature in a black body that absorbs that radiation that is equal to the colour temperature. OK, the real world is not composed of ideal black bodies, but it should at least create a benchmark.
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Batroost, I think you might well find that Kirchoff's radiation law is not an odd thing to say. Feel free to research it.
Lightarrow's point is that if all you see is, say, green light, then you don't know what temperature the object is at. If you have light at 2 wavelengths, say red and blue, then you can asign a temperature to the object (in general the brighter the blue is compared to the red, the hotter the object). What I'm not sure about is how colour temperatures are calculated. I don't think it can be as simple as saying "the object emits more light at 500nm than at any other wavelength. A black body at x Kelvin emits more light at 500nm than at any other wavelength so the object has a colour temperature of x Kelvin."
It would be ridiculous to try to asign colour temperatures to any sort of gas discharge/ fluorescent lamp this way so I guess it's matched by eye (in which case a green laser would be impossible to match to any black body radiation) or by some sort of curve fitting- the spectrum is closest to that of a black body at x Kelvin- again this wouldn't work for a monochromatic light source.
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Batroost, I think you might well find that Kirchoff's radiation law is not an odd thing to say. Feel free to research it.
I suspect that there is some truth in both sides of this.
In principle, I would read this as an extention of the idea in electronics that an antenna that is good for transmitting is equally good at receiving (and visa versa).
It assumes that the processes involved in transmission and absorption of light are indeed reversable.
One area where that would seem dubious is where one is talking about fluorescence, where the absorption spectra may be different from the emission spectra. An extention of this is the famed 'greenhouse effect', where radiation emitted back out into space is of a different spectra from that being absorbed, and consequently an atmosphere that is inert to the incoming radiation can block outgoing radiation, thus allowing a buildup of temperature to a higher level than would be otherwise the case.
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The greenhouse effect is about the transmission of the atmosphere and the emission from the earth's surface. These are not the same object so the law doesn't apply.
it seems that the proof of the law has been debated at some length.
http://www.mzwtg.mwn.de/arbeitspapiere/Schirrmacher_2001_1.pdf
"It assumes that the processes involved in transmission and absorption of light are indeed reversable."
Fair enough, they are. a photon goes in and promotes an atom to an excited state or that excited state decays and gives the photon back. Of course, if the excited state changes then it is not strictly the same body so, for example, if it loses energy to thermal vibration, it no longer exists to have it's absorbtion or emission measured.
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Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.
This is certainly true, but I am not sure how it relates to the argument.
That's very strange, because it was you to make the original question about a laser beam's temperature, to which I have answered!
We say that the surface of the Sun is approx 6000K, because that is the temperature of the light emanating from the surface. We know that the temperature beneath the surface can reach millions of K, but that is not the radiation we receive, so we know that the temperature on the surface is not as hot an the depths of the Sun are.
Yes, you can put a filter in-front of an object, and that will cause the system as a whole to cool on its exterior (beyond that filter), but it does not prevent the interior of the system from being hotter.
The question is, should not the radiation emitted from a 6000K body should be able to induce a temperature in a black body that absorbs that radiation that is equal to the colour temperature. OK, the real world is not composed of ideal black bodies, but it should at least create a benchmark.
Actually I haven't understood what you mean; however, about the max temperature it's possible to achieve concentrating sun's light with lenses or mirrors, I would agree with you (indeed I had never said you were wrong about it); I think we cannot change the sun's spectrum shape (and so, the radiation's temperature) in that way. Not sure, however. Interesting question.
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On a sunny day, i can burn paper with my magnifying glasses. If i had a really good glass, could i burn through steel? or maybe the earths core!!!
What a bright idea! " Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. "Archimedes once said. But does the long lever exist? In the same way, i am afraid that we can not find such a magnifying glass big enough to burn the earth. However, many children often burn ants with magnifier in sunny days. [:o]
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The power available depends on the area of the lens and temperature depends on the inverse of the f number, the problem of obtaining a sufficiently small f number can be overcome by using a two stage focusing system.
The melting of steel has been demonstrated both by this method and by mirrors.
References to follow when I hunt them down.
Whether or not 6000°K can be exceeded is an open question I would like to see comments.
http://gizmodo.com/5069043/solar-furnace-melts-steel-our-minds
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Why would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?
Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.
Lightarrow?
If I get it right you need a whole spectrum to make a decided temperature.
But if it comes to a lasers light, how would you go about to define a temperature for it, isn't it the wavelength/frequency you would use then?
They have an energy, they must have a temperature?
What am I missing?
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Or should I ask about the heat instead?
"Temperature is a number that is related to the average kinetic energy of the molecules of a substance. If temperature is measured in Kelvin degrees, then this number is directly proportional to the average kinetic energy of the molecules.
Heat is a measurement of the total energy in a substance. That total energy is made up of not only of the kinetic energies of the molecules of the substance, but total energy is also made up of the potential energies of the molecules."
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AWhy would a monochromatic LASER not have a colour temperature (it may not have a black body spectrum, but the light still has an energy associated with its frequency, and that energy equates to a temperature)?
Just because it doesn't have a blackbody spectrum. I can find a specific wavelength, let's say 550 nm, in the spectrum of a blackbody of any temperature. I can take a tungsten block at 20°C or at 3000°C, put a very good colour filter in front of it, which let pass only 550 nm light, and how could we establish this light comes from a 20°C or 3000°C block of metal? Certainly at 20°C the intensity of radiation at 550 is less, but this could also be ascribed to source's distance and surphace properties.
It's only the spectrum's shape (specifically: its maximum's position) which is related to temperature.
Lightarrow?
If I get it right you need a whole spectrum to make a decided temperature.
Not only this, you have to be sure that your spectrum is part of a blackbody spectrum.
But if it comes to a lasers light, how would you go about to define a temperature for it, isn't it the wavelength/frequency you would use then?
Absolutely not. As I wrote (3 years ago!) in the post you quoted, you could have two equal light beams, the same intensity, the same wavelenght and coherence, everything the same, but one coming from your laser and the other coming (after adequate filtering equipments) from a big bulb lamp. How do you establish the temperature?
They have an energy, they must have a temperature?
What am I missing?
To define the temperature of an EM radiation you must have a photon gas with a blackbody spectrum. Energy and temperature are two completely different concepts.
Or should I ask about the heat instead?
"Temperature is a number that is related to the average kinetic energy of the molecules of a substance. If temperature is measured in Kelvin degrees, then this number is directly proportional to the average kinetic energy of the molecules.
Heat is a measurement of the total energy in a substance.
No, heat is a way to transfer energy between two bodies: through a difference in their temperature.
That total energy is made up of not only of the kinetic energies of the molecules of the substance, but total energy is also made up of the potential energies of the molecules."
...at least. [:)] There is also another form of energy: mass of particles.
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Thanks LA, one mystery less :)
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Everyone: Thank you for your contributions on this topic. I think it's clear that solar energy can be concentrated sufficiently to melt steel, and a lot of other things too. In the presence of oxygen, steel will burn at those temperatures.
I think this horse has been officially flogged and it's time to move on. Does anyone have violent objections to locking or abandoning this thread?
Of course, I think there are some very important questions and concepts associated with the original question, but it might be more productive if we were to initiate new topics that focus on those specific points.
Wadyafink?
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The melting of steel has been demonstrated both by this method and by mirrors
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http://gizmodo.com/5069043/solar-furnace-melts-steel-our-minds
In this video we can see a metal plate which is melting in a circular region and producing a hole. But why it doesn't become red, then orange, then yellow, then white? Is it steel or tin.... [:)]
(Then the image change, then it comes again on the melted plate with the hole, and this time it's orange-hot...strange [:)]).
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My guess is that the filters they must have used in front of the camera (to stop the incredible brightness washing out the image) must have distorted the colour. Afterwards they show a conventional image as the metal cools down and you see it's still red hot.
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It could be, but it doesn't convince me a lot. Why we don't see any smoke? If there were put filters for brightness, why we don't see the various colors (red, orange, yellow) in the different regions of the plate, even if the entire plate is perfectly visible?
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Are you thinking coherent (lasers) light versus incoherent (sun) light Lightarrow?
Could it be that the effect is so fast, the heat so instantaneous that the atoms around it doesn't have the time to react?
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The melting takes a few seconds so it can't be that.
It's a good question. The molten metal doesn't seem to be glowing as it melts; and it should be.
I still think it's glowing, but you can't see it because of the glare of the multiplied Sun. We normally think of molten steel as bright but, in this context, it's quite dull compared to the incident light.
The same dark glass filters that protect the camera are hiding the glow.
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My link is so phreaking (G3) slow so i usually stay away from movies, but I guess i will need to see this one. Just guessing, I think you're right though Bored Chemist, If it takes seconds there is no possibility of the atoms not being able to react and then it seems as it must have something to do with the light caught in the camera.
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Very cool effect and over quite a large area too, I was expecting it to be more like a laser, a needle sort of :) But I agree, I think it's the camera that gets 'overloaded' by the light even with the filter on, or possibly that it gets melted so quickly and over such an large area that the redness sort of gets hidden under it and . . Awh :) But now that I've seen it I got to admit that I don't know, you can retract the imagery and compare the plates as they melt and after and the plates color seems almost the same even though that the 'after' should be without any Sun-beam working?
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How could we explain the absence of smokes?
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It heats up so rapidly there is little time for any oxidation. The steel is melting rather than burning. When you cut steel with oxygen, you are burning it in an exothermic reaction.
BTW - you can take a trip to France to see one in action (if it's still in operation.)
http://en.wikipedia.org/wiki/Solar_furnace
Edit: Come to think of it, the answer to the original question is, technically, no.
Unless you grind up the steel into small particles, it won't burn. A steel plate can only be burned in an atmosphere that is very rich in oxygen.
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How could we explain the absence of smokes?
Because the steel is only just melting,not burning.
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How could we explain the absence of smokes?
Because the steel is only just melting,not burning.
There's no oxygen in the atmosphere there? [:)] At ~ 1400°C Iron, carbon, iron carbides do burns and generate smoke, IMHO.
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It heats up so rapidly there is little time for any oxidation. The steel is melting rather than burning. When you cut steel with oxygen, you are burning it in an exothermic reaction.
BTW - you can take a trip to France to see one in action (if it's still in operation.)
http://en.wikipedia.org/wiki/Solar_furnace
Edit: Come to think of it, the answer to the original question is, technically, no.
Unless you grind up the steel into small particles, it won't burn. A steel plate can only be burned in an atmosphere that is very rich in oxygen.
It doesn't have to burn completely to generate smoke, only a little part.
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It heats up so rapidly there is little time for any oxidation. The steel is melting rather than burning. When you cut steel with oxygen, you are burning it in an exothermic reaction.
BTW - you can take a trip to France to see one in action (if it's still in operation.)
http://en.wikipedia.org/wiki/Solar_furnace
Edit: Come to think of it, the answer to the original question is, technically, no.
Unless you grind up the steel into small particles, it won't burn. A steel plate can only be burned in an atmosphere that is very rich in oxygen.
It doesn't have to burn completely to generate smoke, only a little part.
I'm sure there is a small amount of smoke produced, but at that temperature any carbon particles (soot) will be converted into CO2 rather quickly.
BTW, despite the fact that this is quite an impressive demonstration of localized high temperatures, it's not really a very good demonstration of power production. I can burn a hole in a chunk of steel almost as quickly with my oxyacetylene cutter.
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So the answer is, not really with a lens, but yes with parabolic/concentrating mirrors. This article cited earlier in the thread talks of making experimental alloys and 6,000F http://www.time.com/time/magazine/article/0,9171,909204-1,00.html so certainly a steel sheet is not a problem. And yeah, smoke is from incompletely burned materials, so here they're raising it to its melting temperature, not 'burning' it with oxidation (not intentionally), like they'd do with a graphite electrode in a smelting crucible. Pours right out.
I'm still confused about the whole temperature / heating / light wavelength back-discussion.
I've wondered this for several years, and I just found out:
http://www.3drender.com/glossary/colortemp.htm
My friend works in film and they're always concerned about their light source - Keno Flos, natural sunlight, filters, 10Ks, blah blah - and apparently it dates back to William Kelvin's observations of a heated block of carbon. It has nothing to do with the heat of the filament or radiant body's burning. And so now I get what people meant by "black body" - not as in a non-visible light radiation spectra, but literally a black-bodied chunk of graphite.
The video is of an experimental parabolic concentrator, not a power-generating one; for a power generating one, they only heat a boiler as much as they want the steam temperature to be at for their turbine otherwise they'd melt the tower..
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I'm still confused about the whole temperature / heating / light wavelength back-discussion.
The concept of "Temperature" is not actually as simple as it would seem.
Temperature is a statistical concept, you can't define temperature for a single particle and not even for some. In this case the particles are photons; you must have a lot of them , and so a lot of frequencies, to define T.
In a recent discussion I had in another forum, a physics professor wrote that when a physical system A exchanges energy with another system B only by means of a perfectly monocromatic laser beam, it doesn't exchange heat...