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Radiation View Factors

Assume diffuse, grey surfaces with uniform radiosity.

The heat transferred by radiation from surface1 to surface 2 is:

q12=sigma*(T1^4-T2^4)/((1-e1)/e1/A1+1/A1/F12+(1-e2)/e2/A2)

Question 1.

Surface 1 is the bottom surface of a finite rectangular area sitting above (and parallel to) an infinite plane (surface 2). The view factor, F12, is

1.0 in this case. The problem I see here is that A2 is infinite. This doesn't cause any real mathematical problems in the equation above, but it does seem to mean that the emissivity of the infinite plane is no longer relevant--it is as if the infinite plane has emissivity of 1.0, regardless of its actual emissivity. This doesn't "feel" right to me. Does it make sense? Does having infinite space to absorb heat mean that all heat will be absorbed, even when the emissivity is less than 1.0?

Question 2.

There are two rectangular areas in parallel planes. The directions of the rectangles are aligned with each other. One rectangle is not necessarily directly above the other. Is there a general expression for the view factor between the two rectangles? I cannot make any simplifying assumptions. (i.e. I cannot say that one area is very small; I cannot say that the vertical spacing is very large or very small relative to anything, etc.) I know there is a simple solution if the two areas are very small relative to the distance between them. I have also found (in

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a somewhat more general solution (where one rectangle is very small relative to the other, and located directly below one corner of the larger area). The source also seems to imply that this can be expanded into the general case through algebra, but then it uses notation I can't understand (and then says the results may not be accurate!) Can anyone help, or am I stuck doing numerical integrations?

Thanks,

-Paul

Reply to
Paul Skoczylas
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Dear Paul Skoczylas:

If this is homework, you should find published "shape factors", with various geometries in your book...

David A. Smith

Reply to
N:dlzc D:aol T:com (dlzc)

  1. It's not homework.
  2. A shape factor is not the same thing as a view factor.
  3. The view factor I want is not one of the ones in my book. It's not even directly given in that technical note (containing many view factors) I referred to.

-Paul

Reply to
Paul Skoczylas

The view factor will be different for each portion of the infinite plane. Maybe it would help to think of it as a coplanar set of tiny, adjacent rectangles, each with its own view factor W.R.T. finite surface number 1. The resultant radiation and absorbtion will be an integral that resolves an infinite number of different view factors into a single expression.

You're stuck with integrations in any case, I think. Even when two identical rectangles are facing each other, and perfectly aligned, the radiation/absorbtion aren't as simple as a bunch of straight lines perpendicular to the surfaces. Heat transfered from the southwest corner of rectangle 1 to the northeast corner of rectangle 2 will travel farther, and will be less intense, and will have a different angle of incidence, than heat traveling the shortest (perpendicualr) distance between the surfaces.

Every point on one rectangle radiates heat that any point on the other rectangle can absorb. Even the simplest situation has to take this into account, and even the most simplified expressions for radiation/absorbtion have to deal with an infinite number of points, distance, angles, etc., if they're to be accurate. The only reason there's a simple expression when two rectangles are small relative to the distance between them is that the angles/distances between any pair of points are all pretty much the same. The only time such a simplification could be totally accurate would be if the rectangles were infinitely small.

KG

Reply to
Kirk Gordon

Actually, I've found that it can be solved analytically. The solution was in Can. J. Chem. Eng., Vol 45, Feb, 1967, by Chia-Jung Hsu.

The equation takes up two thirds of a page, but it seems to work (i.e., it matches the results from more specific [simpler] cases for which other results are available, and matches the approximate result in situations when the approximate result can be expected to be accurate [rectangles small relative to the distance between them]).

Reply to
Paul Skoczylas

I'll retract that one point from my earlier message. Apparently what I learned as "view factors" are referred to in some sources by either "shape factors" or "configuration factors". My text uses "Shape Factor" in a completely different context; other texts vary on usage. Forgive my error there.

I did find the equation I was looking for. It certainly isn't in any basic heat transfer textbook. Even a book I found specifcally about radiation heat transfer didn't have it. Thankfully it did have a "catalogue" of various factors, which told me where I could find it!

-Paul

Reply to
Paul Skoczylas

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