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OK energy gurus, I have a question - 5/12/2011 5:19:29 PM   
Termyn8or


Posts: 18681
Joined: 11/12/2005
Status: offline 		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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RE: OK energy gurus, I have a question - 5/12/2011 6:32:52 PM   
TheBanshee


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Status: offline 		I just had a flashback from my Algebra class

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RE: OK energy gurus, I have a question - 5/12/2011 6:58:01 PM   
pahunkboy


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Joined: 2/26/2006
From: Central Pennsylvania
Status: offline 		I doubt you will melt metal.


You could tho try to get a fan blade to go round over it-  and save money on your electric bill.

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RE: OK energy gurus, I have a question - 5/12/2011 7:02:56 PM   
Termyn8or


Posts: 18681
Joined: 11/12/2005
Status: offline 		Well removing convection and most of conduction there is only the radiation of heat. BTUs and caloric measurements of heat do not take into account absolute temperature. I could've said in a vacuum like in outer space, but that raw figure would require quite a bit more process before it got to what I can get on Earth.

Actually even that figure (in space) would be useful, and would be a part of it. But I only tried to simplify it, not make it into a grade school question. Regardless, I don't have that absolute figure either. Even that would help.

T^T

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RE: OK energy gurus, I have a question - 5/12/2011 7:04:08 PM   
Termyn8or


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Status: offline 		"I doubt you will melt metal. "

I think I would, but just what kinds of metal can I melt ?

T^T

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RE: OK energy gurus, I have a question - 5/12/2011 7:05:05 PM   
dcnovice


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quote:

ORIGINAL: TheBanshee

I just had a flashback from my Algebra class


The train reaches New Haven at 12:45.

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it's never enough to keep up.

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RE: OK energy gurus, I have a question - 5/12/2011 7:14:24 PM   
samboct


Posts: 1817
Joined: 1/17/2007
Status: offline 		Hi Termy

I think the number you need to start with is that there's about a kW (1,000 watts) per square meter of sunlight. So figure out the size of your fresnel lens in meters and what it focuses down to and that will get you watts/cm. But then you run into the problem of absorption- how much light does whatever you're trying to cook adsorb versus reflect? This is a problem. But to give you some idea of the orders of magnitude- a household iron uses about 200 watts or so IIRC and that's heating something around 20 sq. inches. Assume it's linear and an iron can go to what- 300C? So you might have more watts going into a smaller area-but I think the problem is actually getting those watts into that area.

Sam

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RE: OK energy gurus, I have a question - 5/12/2011 7:15:51 PM   
Hillwilliam


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Status: offline 		I just had a wierd thought about that lens and an anthill.

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RE: OK energy gurus, I have a question - 5/12/2011 7:34:26 PM   
Icarys


Posts: 5757
Status: offline
quote:

I think I would, but just what kinds of metal can I melt ?

About any type that would melt under 2000F. "Most" large lenses pump out that much. Who knows, you may be able to compound them in some way. I'm not sure if this would work or not but try placing one above the other. making sure though that it isn't focused too much. You'll melt the bottom lenses lol.

Just get the damn lenses out and burn some shit already. Wear sunglasses!

< Message edited by Icarys -- 5/12/2011 7:47:15 PM >

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RE: OK energy gurus, I have a question - 5/12/2011 9:37:30 PM   
TheBanshee


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quote:

ORIGINAL: dcnovice


quote:

ORIGINAL: TheBanshee

I just had a flashback from my Algebra class


The train reaches New Haven at 12:45.


and the other leaves Chicago going at  60 miles per hour...

(in reply to dcnovice)
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RE: OK energy gurus, I have a question - 5/12/2011 11:20:48 PM   
Termyn8or


Posts: 18681
Joined: 11/12/2005
Status: offline 		Let's get the show on the road
At the curtain take a bow
New Haven just a rendevous to take you to
A lover who was then but never now

And Susan paid the lady
Who called the players' song
Just a figurine of stagehand reveries
I guess up-and-coming can't be that wrong

Let's get the show on the road, babe
The spotlight's on the stage
Somehow it seems that I heard these words before
Did you forget to turn the page?

Therefore the answer is "spirogyra".

T^T

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RE: OK energy gurus, I have a question - 5/13/2011 12:24:07 AM   
pyroaquatic


Posts: 1535
Joined: 12/4/2006
From: Pyroaquatica
Status: offline 		Large Crystal Clear Fresnel Lenses can melt metal. Granted it cannot melt large blocks of metal because energy dissipation. So you would have to stick with filings or shavings. Maybe a few nuts and bolts.

Busting open Projection Televisions Term? 

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As your desire is, so is your will.
As your will is, so is your deed.
As your deed is, so is your destiny.
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RE: OK energy gurus, I have a question - 5/13/2011 3:44:02 AM   
shallowdeep


Posts: 343
Joined: 9/1/2006
From: California
Status: offline 		Because I'm procrastinating and haven't used my thermodynamics textbook in awhile…

Rough upper bound:

Assume emissivity and absorptivity of the heated material are equal. Assume a perfectly insulated material with no conduction or convection, so all heat transfer will occur as radiation.

Stefan-Boltzmann constant (sigma): 5.67 x 10^-8 W/(m^2*K^4)
Area of the material's surface exposed to radiate: 4 in^2 = 0.00258 m^2
Area of the lens = 1200 in^2 = 0.7742 m^2
Incident radiated heat energy transfer rate (Q_dot_incident): 1000 W/m^2 * area of lens = 774 W (no losses)
T_s = absolute temperature of the surface (in Kelvin)

The material will radiate heat with rate:
Q_dot_emit = emissivity * sigma * area of material * T_s^4

And absorb heat with rate:
Q_dot_absorb = absorptivity * Q_dot_incident

Equating the two to solve for a final, steady state temperature:
T_s = (Q_dot_incident/(sigma * area of material))^.25
= (774/(5.67E-8 * 0.00258))^.25
= 1517 K = 1244 ºC = 2270 ºF

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RE: OK energy gurus, I have a question - 5/13/2011 3:50:13 AM   
sunshinemiss


Posts: 17673
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Sunny
Quote of the Day
goes to
shallowdeep

for


Assume emissivity and absorptivity of the heated material are equal. Assume a perfectly insulated material with no conduction or convection, so all heat transfer will occur as radiation.

Stefan-Boltzmann constant (sigma): 5.67 x 10^-8 W/(m^2*K^4)
Area of the material's surface exposed to radiate: 4 in^2 = 0.00258 m^2
Area of the lens = 1200 in^2 = 0.7742 m^2
Incident radiated heat energy transfer rate (Q_dot_incident): 1000 W/m^2 * area of lens = 774 W (no losses)
T_s = absolute temperature of the surface (in Kelvin)

The material will radiate heat with rate:
Q_dot_emit = emissivity * sigma * area of material * T_s^4

And absorb heat with rate:
Q_dot_absorb = absorptivity * Q_dot_incident

Equating the two to solve for a final, steady state temperature:
T_s = (Q_dot_incident/(sigma * area of material))^.25
= (774/(5.67E-8 * 0.00258))^.25
= 1517 K = 1244 ºC = 2270 ºF





http://www.collarchat.com/m_3671060/mpage_1/key_/tm.htm#3671798

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RE: OK energy gurus, I have a question - 5/13/2011 4:01:25 AM   
mummyman321


Posts: 2102
Joined: 10/31/2005
From: Dusseldorf
Status: offline
quote:

ORIGINAL: pyroaquatic

Large Crystal Clear Fresnel Lenses can melt metal. Granted it cannot melt large blocks of metal because energy dissipation. So you would have to stick with filings or shavings. Maybe a few nuts and bolts.

Busting open Projection Televisions Term? 


I have melted pennies and quarters using a 24" X 24" Fresnel lens before.

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Life - Its not about where you are but about the journey to get there - I prefer to choose the road less traveled

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RE: OK energy gurus, I have a question - 5/13/2011 5:38:37 AM   
Icarys


Posts: 5757
Status: offline 		My guess is, you want to melt the steel and cast something?

_____________________________

submission - the feeling of patient, submissive humbleness - the state of being submissive or compliant; meekness.

Alaska Bound-The Official Countdown Has Started!
http://tinyurl.com/872mcu3
http://alturl.com/mog7m

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RE: OK energy gurus, I have a question - 5/13/2011 10:51:43 AM   
Icarys


Posts: 5757
Status: offline 		Try this.

Damn that's hot.

< Message edited by Icarys -- 5/13/2011 11:33:00 AM >

_____________________________

submission - the feeling of patient, submissive humbleness - the state of being submissive or compliant; meekness.

Alaska Bound-The Official Countdown Has Started!
http://tinyurl.com/872mcu3
http://alturl.com/mog7m

(in reply to Icarys)
Profile   Post #: 17
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