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Termyn8or -> OK energy gurus, I have a question (5/12/2011 5:19:29 PM)
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This will take a bit of math. I have access to a bunch of rather large fresnel lenses. The latest one is 50" diagonal with a 3:4 aspect ratio. That means it is 30" by 40", which is 1,200 sq. inches.
We're at about 41 degrees latittude here, so let's call it 45 to make it easy. Now I know there is a figure for wattage, BTUs, calories or whatever heat energy from the sun. So figure at the equinoxes we're about halfway up from the equator. All that can be figured out. But that's not the figure I'm after.
Now this 1,200 sq. inches of lens will typically ocncentrate down to about 4 or 5 sq. inches at the proper distance from an object. Though the optimum focal point may be different for heat slightly, it's not all that far off. The actual figure I seek is what is the maximum possible temperature I can achieve.
Now the amount of energy concentrated might actually vary due to several factors, a lower amount will heat something up more slowly of course. However discarding disspation by air currents and such, just how hot can I get something ? I can get rid of the wind by building an enclosure. I can thermally isolate the object to be heated as well (except from still air), so discarding those factors for now, in still air at normal pressure, what kind of thermal energy is on tap here ?
I have burned asphalt with one. So I'm sure I could melt tin or lead. What I want to know is the highest temperature attainable by this method. Can I melt steel ? Can I melt aluminum ? Can I melt nickel ? Can I melt ?, well you know what I mean. Things have melting points and no matter how much heat energy you can throw at it, if you can't attain the temperature all you have is a very hot solid.
For figures just use the mean. Average relative humidity because I don't intend to pull a vacuum. We are not much above sea level so basically normal barometric pressure. Average humidity is a bit high here, but not quite tropical.
It's like it's very difficult to cut steel with a propane torch because it can't deliver enough heat. That's why they use acetylene. I'm just wondering if there is a way to figure this out without resorting to actually measuring the temperature, like with a thermoelectric pile, which I think would be just about the only way to do it. It might boil mercury, but I'm not sure. Even if I knew the boiling point of mercury offhand that would do me no good.
An ideas here or is this something with too many variables ? I see the variables as :
1,200 square inches reduced to three or four square inches.
Trasmission can be disregarded for now.
Dissipation is pretty much normal based on atmospheric pressure and humidity.
Assuming 45 degrees latitude at equinox for simplicity's sake.
Assuming that I am not trying to melt a blackbody, and I can figure out the proper focal point for heat rather than light.
It sounds simple at first, but after thinking about it, it's not so simple. That's why I ask in a BDSM forum LOL.
T^T
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