heat transfer by radiation 2

[macmet]

I have searched this site, my texts, and the internet looking for an answer to my question which is based on the fundamentals of heat transfer by radation. I'm sure I'm overlooking something simple here, but I cannot figure out what. This is not a school question, I have recently applied to a job that requested a basic understanding of heat transfer by radiation and I feel I should be able to answer this anyway.

My question relating to heat transfer radiation has to deal with the distance between two bodies.... I've gone through all my old texts and looked at websites and I have yet to see a comment about effects of varying distances on the radiation transfer other than it is irrelevant.

Can someone explain to me then how when standing beside a fire , say 1 ft away, I can feel it intensely but if I were to step back 10 feet, I would feel warm, but nothing more. If I were to walk back 100 feet, I wouldn't feel it at all?

The only thing I can guess is that at 1ft, I am getting the radiation effects plus conduction (heat transfer through the air). At 10 ft, I am not longer experiencing conduction, so I only feel radiation after passing through cool air, which acts as an insulator. And at 100 ft, the insulating effect of the air has totally eliminated the radiation effect.

But, if tha'ts true, shouldn't it matter what the medium is as well?

Cheers

 

[MintJulep]

Inverse square law.

http://en.wikipedia.org/wiki/Inverse-square_law

 

[chicopee]

Well, make a sketch. Assume the fire to be a small sphere, lets say two feet in diameter. Then draw three full circles(representing spheres) of radii 1',10' and 100' with a common center occupied by the 2' dia. sphere. Along the horizontal axis draw (symmetrically 3' on both side of the x axis)a 6' line ( representing a person) at the 1',10' and 100' position. Then draw lines extending from the center of the fire to the head and feet of the person. By now you get the idea that the incident radiation on the person becomes less as the distance from the fire increases.

To prove it mathematically is a little bit more difficult but assume constant radiation density radially emitted from the small sphere, the amount of radiation that you will receive will be a ratio of the solid angle subtended by the person's body to that of a sphere at those intervals of 1', 10' and 100'. Note radiation intensities will decrease as the inverse of the radius square,ie, at the 1',10' and 100'.

 

[macmet]

Well thanks for the quick responses. I think my problem is that whenever I go through an example it seems to be regarding radiation from the sun. And the flux from teh sun is always the same. So that it's not so much that the distance doesn't matter in these examples I see, it's that the distance is constant.

I think I see where you were going with that example chicopee, thanks.

Mintjulep, I checked out that link and it was helpful too. Thanks.

 

[IRstuff]

A point radiative source, like the sun, or the fire, has a constant flux per unit solid angle. The solid angle is defined by the physical area of interest, your body, divided by the distance squared. Therefore, you get only 1% of the flux at 10 ft that you got at 1 ft, which is the basis of the inverse square law.

If you look at your heat transfer text, a free one is available here (

http://web.mit.edu/lienhard/

http://www/ahtt.html),

you'll see that conduction becomes negligible beyond about 1/2 inch, so even at 1 ft, it's all radiation.

If you are ever at a Disney resort, you can verify this by the fact that the gas flame bursts shower you with heat way quicker than the speed of sound.

TTFN

FAQ731-376

 

[davefitz]

The value to look up in the texts is the "view factor" F. One simple method of calculating the view factor is called the " Hottels crossed string method".

View factors for typical orientations and bodies are tabulated in some texts, such as Siegel + Howell's "thremal radiation heat transfer" or Mike Modest's text.

 

[JStephen]

Ditto to the View Factor answer above, which is the solution to your question.

 

[IRstuff]

Two additional points.

> If you were to feel constant "heating" regardless of distance, you would need an infinite energy output, since the implication is that if you were a million miles away, you could still feel the heat. That's patently absurd, and there is only one type of source that can even come close to doing that, and if it did come our way, all life on Earth would instantly cease to exist.

> A radiative heat source is no different than a 60W light bulb, which allows you to do reading close up, but is pretty useless as a light from 100 ft away.

TTFN

FAQ731-376

 

[macmet]

Well thanks everyone again for the tips. I am reading up on view factors right now because of it.

Cheers.

 

[racookpe1978]

Equally important - as a "basic law" of radiation heat transfer are several subtle "rules" of the radiation heat exchange:

BOTH bodies in the problem are radiating "heat energy" = you (the colder one) are not radiating as much, but you ARE radiating heat away from yourself in all directions, not just absorbing it from the hot body. The "hot body" is receiving a portion of that radiating heat form you. The "walls" of the chamber you are in are receiving radiation form you, the hot body, and the rest of the walls. Scratch all of those problems IF and ONLY IF you are in a prefect vacuum of infinite size. Space, in other words. Otherwise, just mention it as a complication you have to make adjustments for in the real world.)

The amount of radiation being lost is proportional to the body's surface (metal) condition and roughness and color and reflectivity. A "theoretical" black body is impossible to get in real life, so dirt, shininess, shape and texture matter. A LOT. In a theoretical problem, they can be ignored - but again, only if you make that statement that you are (deliberately) ignoring the complications. (This implies, of course, that "you" know how to take care of the complications in your real life of real work and real materials .....)

Radiation is released/emitted according to the 4th power of the ABSOLUTE temperature (degrees K - NOT ever degrees F or degrees C.) Get familiar with the conversion, and KNOW IT. Know why you need to use degrees K. Know room temperature, freezing temperature, and boiling temperature in degrees K - and memorize them.

 

[ginsoakedboy]

Stars for davefitz and racookpe1978.

Macmet, also try to understand the relationship between transmissivity, reflectivity and absorptivity. In addition, the wavelength dependence of all of these properties.

 

[corus]

As another question,
racookpe1978 refers to the color (or colour in the UK) but what is the colour you see? If you heat a metal bar until it glows red then is it still black but only appears to be red because of the radiation you see? Just a thought.

corus

 

[ione]

Radiation is an intended as the electromagnetic wave of a “hot” source, where hot denotes a temperature above the absolute zero (0 degrees Kelvin).

From the Wien’s displacement law:

T*?max = b

where:
b= 2.8977685×10?3 mK Wien’s displacement contant
T = Absolute temperature of the emitting source [degrees Kelvin]
?max = peack wave length [m]

As temperature increases the peak emissivity shifts towards shorter wave length. We can deduce that as temperature varies the color (or colour in the UK) changes.