## Radiation view factor confusion

## Radiation view factor confusion

(OP)

So I have consulted multiple sources both in books and online regarding the various view factor geometries and their formulae but I've yet to find one where the outer shell of a hemisphere (convex surface facing up) is situated in the center of a square cavity and exposed to the inside surfaces of that cavity. Is there a formula for that? I would think if you have a hemisphere in the middle of a box, the view factor would be a little more than 50% because some parts of the hemisphere would "see" more than half the box over its horizon but this is just an intuitive guess. I can't find the formula anywhere. Any help would be much appreciated. Thank you!

## RE: Radiation view factor confusion

TTFN (ta ta for now)

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## RE: Radiation view factor confusion

## RE: Radiation view factor confusion

The usual approach is to mathematically describe the view factor and integrate over the supposed field of view. The following is from Radiative Heat Transfer by Modest. Note the cosine terms, which basically zero out the radiative transfer for incidence angles at 90 deg from the normal of the receiving surface differential area. Anything below that angle is essentially invisible to that differential area

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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## RE: Radiation view factor confusion

## RE: Radiation view factor confusion

"...when logic, and proportion, have fallen, sloppy dead..." Grace Slick

## RE: Radiation view factor confusion

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

## RE: Radiation view factor confusion

Then I found this video: https://www.youtube.com/watch?v=9cTE7pdlnHg

, which circumvents having to do any calculus whatsoever because 1 of the terms cancels out. In that video he uses a cylinder. Since all surfaces of the cylinder have divergent normals, the cylinder can't see itself and F22 = 0. This means F21 = 1 and via reciprocity you can find F12. In my case I'm using a hemisphere inside of a cube rather than another sphere so I have to make 2 assumptions:

Since a hemisphere also has divergent normals, F22 should also be 0, leading to the same conclusion as the video, that F12 is merely a ratio of areas between 1 and 2.

Via Nusselt, the "patch" projected by a differential area from 2 onto 1 can be instead projected onto a virtual sphere and should cover the same proportional area of the enclosure. So imagining the oven as a sphere instead of a cube, may change the patch shape and total area but shouldn't much change the patch's proportional area.

From these assumptions, A1F12 = A2F21 or F12 =A2/A1 (F21 = 1). Therefore F12 = 0.5 (4piR^2)/(4piR^2). As it turns out, this is a very very small number (0.07) for a 12 lb turkey, giving me an unsatisfactory contribution from radiation.

I am already assuming the metal surfaces are perfect black body radiators and the turkey has emmissivity of 0.8 (it's been shown that meat's emmissivity drops as it browns). I can add a fudge factor to the view factor because the meat cannot be idealized as a sphere but will always have more area than that perfect shape, but this might only buy me another 10-15% or something. I still feel like either I made a mistake or there's something missing from the radiation side.

## RE: Radiation view factor confusion

## RE: Radiation view factor confusion

I would further suggest that given where you are in your understanding would be to assume that the oven is a concentric hemisphere, which would make the solution much simpler. You can then refine from there

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Radiation view factor confusion

IRstuff: This video is also good, perhaps better than the more simplistic one I found but I don't see anything particularly new there that explains the document you and I both referenced. I think I more or less grasp the concept of using simple rules (like symmetry, reciprocity, etc) to make a complex problem more simple to solve. In my example, if I take the oven to be a sphere and the meat to be a sphere, then there are only 2 surfaces to consider in which case I assume I went about the problem correctly in my above post (ok, if you remove the 0.5 part and consider the meat a whole sphere)? If I insist on making the oven a box, the only difference is now I have 7 surfaces (6 from the enclosure and 1 from the meat). This would yield a better answer (slightly) but since all surfaces are planar, even less radiation actually gets to the meat, so this exacerbates the problem (the problem being that I know from real life the radiative contribution is similar to the convective contribution).

Often times textbooks will have meat in oven examples where the total h value is provided to you and it something like 20 or 22. In a common set of initial conditions I have a convective h of about 12 (makes sense) but a radiative h of only 3 and a bit. Total h of 15 isn't that bad but it's a little low. I'm wondering if the highly polished surface of the oven allows rays that miss the meat from 1 wall to rebound and strike the meat anyway but something tells me that this could never be better than a perfect black body, which would simply absorb them and re-emit them with no losses.

It's a stumper.

## RE: Radiation view factor confusion