Does that even matter? The incidence angle is essentially zero, so ostensibly, zero, or negligible, radiative transfer.

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It seems that every point on the hemisphere will have a different view factor.

Quote:

It seems that every point on the hemisphere will have a different view factor.

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

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IRSTUFF: I'm familiar with the piecewise differential approach but only in the conceptual sense of explaining view factor. I've never had to actually use it to integrate some arbitrary geometry. Usually the formulae are provided because of how math intensive it is to compute various common geometries. Attached is a simplified drawing of what I'm trying to calculate. If it looks familiar, I'm modelling a piece of meat (the part which is usually exposed) in the center of an oven. To be honest, considering how highly applicable heat transfer as a science is to ovens as a daily application, I'm quite surprised that every chapter in heat transfer doesn't have an oven/turkey example and that there aren't mathematical models in abundance for it. But this is essentially what I'm trying to model and my inability to do so is screwing up my larger calculations for radiative heat transfer. Just from inspecting my doodle, it seems like "some" part of the meat can see a relatively large span of oven but most of the contribution is still from the top half only, leading me to believe the actual view factor is a bit more than 50%. Maybe 60% or thereabouts. Still it would be nice if I could calculate a better approximation rather than taking a guess. Math is the bottleneck for me in this case. I know calculus but I'm a bit rough with abstract methodologies so I'm afraid I have trouble following general math advice (such as the need for piecewise integration) to a very specific conclusion (a formula with well defined variables) without a bit of hand-holding.

• https://files.engineering.com/getfile.aspx?folder=cc516ebb-f183-4938-ab39-3

For the case of a turkey in an oven it would be tough to provide an exact mathematical solution, but one could numerically solve the issue by breaking up both the turkey and the oven into 100 subelements apiece and let the computer do the hard work integrating the equations. It is equally difficualt to exactly know the temperature of the entire oven's interior surface if the only source of heat is a gas fire ( or electric heating element) under the turkey. The practical problem is mostly solved by using a convective oven with a high convective heat transfer coeficient via a hot circulating air fan.

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

http://webserver.dmt.upm.es/~isidoro/tc3/Radiation...

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I found that URL a while ago and was mulling over it yesterday. The closest approximation I could find was at the bottom of page 14 but they never say what "the enclosure" is and they only give you F13, not F31.

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.

A sketch of the geometries that you described in your OP would help. Nonetheless I got a couple of reference that may help you which I found on the webb. 1) A catalog of Radiative Heat transfer 2)Isodoro Martinez "Radiative View Factors. Also, Study Hottel Crossed Strings which is used to determine view factors by plotting to scale the geometries under study.

I suggest you watch THIS video https://www.youtube.com/watch?v=4OqVQlnQ__c&t=... Your comments about F13 and F31 show that you didn't absorb a key concept.

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

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Chicopee: Thanks... and I have attached a sketch in post 5 above.

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.

See fig 5-16a for F21 in Perry Chem Engg Handbook -7th edn - that's the nearest approximation I can find on this.