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Two aligned 1 × 2 m plates 1 m apart: Hamilton–Morgan view factor (NACA TN 2836 Table 12) and the grey-body network with re-radiating surroundings

Tool	Radiation View Factors
Source	Hamilton, D. C., and Morgan, W. R. (1952). Radiant-Interchange Configuration Factors. NACA Technical Note 2836, National Advisory Committee for Aeronautics, Washington, DC. Appendix C, configuration A-1, and Table 12; Oppenheim, A. K. (1956). "Radiation Analysis by the Network Method." Transactions of the ASME, Vol. 78, No. 4, pp. 725–735.
Level	Undergraduate

Problem

Two identical 1 m × 2 m rectangles face each other 1 m apart, aligned. The lower one is at 500 K with emissivity 0.8, the upper at 300 K with emissivity 0.5; the surroundings beyond them are re-radiating. Find the view factor from the Hamilton–Morgan closed form, check it by reciprocity and the summation rule, assemble the radiosity network and find the net radiative heat rate of each surface, the radiosities and the temperature the re-radiating surroundings take; compare with the two-surface formula.

Inputs to Enter

Configuration A-1 (aligned parallel rectangles): a = 1 m, b = 2 m, c = 1 m
T_1 = 500 K, ε_1 = 0.8; T_2 = 300 K, ε_2 = 0.5 (absolute scale); surroundings re-radiating; no shields

Open the Radiation View Factors tool (loaded with this example)

Answers

x = b/c = 2, y = a/c = 1: F_12 = 2/(π x y){ln[(1 + x²)(1 + y²)/(1 + x² + y²)]^½ + y√(1 + x²) tan⁻¹[y/√(1 + x²)] + x√(1 + y²) tan⁻¹[x/√(1 + y²)] − y tan⁻¹y − x tan⁻¹x} = 0.28588 (Table 12 cell x = 2, y = 1: 0.28588; scipy dblquad 0.2858754)
A_1 = A_2 = 2 m², F_21 = F_12 by reciprocity, F_13 = F_23 = 1 − 0.28588 = 0.71412 reach the surroundings
E_b1 = σ 500⁴ = 3,544.0 W/m², E_b2 = 459.30 W/m²; resistances (1 − ε_1)/(ε_1 A_1) = 0.125 m⁻², 1/(A_1 F_12) = 1.7490 m⁻², (1 − ε_2)/(ε_2 A_2) = 0.5 m⁻²
Network with the re-radiating surface 3: J_1 = 3,269.1, J_2 = 1,558.9, J_3 = 2,414.0 W/m²; q_1 = 2,199.1 W leaves the hot plate and all of it reaches the cold plate (q_2 = −2,199.1 W, q_3 = 0); T_3 = (J_3/σ)^¼ = 454.2 K
Two-surface formula (surroundings ignored): q_12 = (3,544.0 − 459.3)/(0.125 + 1.7490 + 0.5) = 1,299.4 W, 41 % less than the network result because it drops the exchange by way of the surroundings

Notes

The tool's default inputs are this case; press Submit. The view factor is one cell of the report's Table 12 (the test script test_radiation_view_factors.php reproduces 80 cells of that table and the P-1, P-2, P-6, A-2, A-5, A-7 and A-8 tables, 711 checks); the network numbers come from an independent numpy solve in References/Radiation/checks/radiation_check.py. Choosing large black surroundings at 300 K instead gives q_1 = 4,772.8 W.