Radiant heat · perpendicular
Radiant heat flux along a wall at 90° to an opening
Our wall runs away from the emitter's plane at a right angle, so openings in it see the emitter sideways. This gives the flux all along the wall, the solid return wall needed before an opening can start, and the effect of a wing wall or an offset.
Result
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For your report
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Working
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Method
The emitter is a uniform rectangle at temperature T with emissivity ε, emitting E = σεT⁴. A receiver on our wall faces sideways towards the emitter. With the receiver at the emitter's mid-height, the emitter splits into equal upper and lower halves of height h = a/2. Each half uses the factor for an element at 90° to a rectangle, level with its corner: φp(X, Y) = 1/(2π) [atan(1/Y) − Y/√(X²+Y²) atan(1/√(X²+Y²))], with Y = c/h.
Configuration factors add over area, so a wing wall is handled by subtraction: φ = 2 [φp((W+b)/h, c/h) − φp(W/h, c/h)]. This is exact when the emitter is on the plane of the wing wall. If the emitter is also set back from the corner (s > 0), a wing wall on the boundary hides more of the emitter than its own length, so the result is conservative.
The solid length L is the distance from the corner beyond which the flux stays at or below the criterion. It is solved exactly and then rounded up. A common slip is to use the formula for a receiver lying flat. Openings are in a vertical wall, so both ratios are divided by the emitter's half-height h. See validation.