Assumes A_1 = A_2 = 1 m² and view factor F_12 = 1 (infinite parallel plates).
Left node = σT_1^4 / right node = σT_2^4 / surface resistances R_1, R_2 and space resistance R_12 connect in series
The radiative heat flow is computed by treating σT^4 as a driving potential and connecting surface and space resistances in series, just like an electric circuit.
Surface resistance (reflection loss when ε < 1):
$$R_s = \frac{1-\varepsilon}{A\,\varepsilon}$$Space resistance (geometric loss controlled by the view factor F_12):
$$R_{12} = \frac{1}{A_1\,F_{12}}$$Heat flow between the two surfaces. σ = 5.670×10⁻⁸ W/(m²·K⁴):
$$Q_{12} = \frac{\sigma\,(T_1^4 - T_2^4)}{R_1 + R_{12} + R_2}$$Blackbody limit (ε_1 = ε_2 = 1) and gray efficiency:
$$Q_{BB} = \sigma A (T_1^4 - T_2^4),\quad \eta = \frac{Q_{12}}{Q_{BB}}$$For infinite parallel plates (F_12 = 1, A_1 = A_2 = A) this simplifies to η = 1/(1/ε_1 + 1/ε_2 - 1).