Radiation between two parallel grey plates

Setup

Two large parallel plates, so that all radiation leaving one reaches the other: $F_{12} = F_{21} = 1$. Plate 1 at $T_1$ ($\varepsilon_1$), plate 2 at $T_2$ ($\varepsilon_2$), $T_1 > T_2$, both opaque grey surfaces ($\rho = 1 - \varepsilon$).

Radiosity $J$ is the total radiation leaving a surface (emitted plus reflected). The net heat leaving a surface:

$$Q = \frac{E_b - J}{(1-\varepsilon)/(\varepsilon A)}$$

where $(1-\varepsilon)/\varepsilon A$ is the surface resistance. The exchange between two surfaces uses the space resistance $1/(A_1F_{12})$.

Network

 Eb1   J1              J2    Eb2
  o--/\/\--o----/\/\----o--/\/\--o
  (1-e1)/   1/(A1F12)   (1-e2)/
   (e1 A1)              (e2 A2)

Total resistance (series):

$$R = \frac{1-\varepsilon_1}{\varepsilon_1A} + \frac{1}{AF_{12}} + \frac{1-\varepsilon_2}{\varepsilon_2A}$$

With $A_1 = A_2 = A$, $F_{12} = 1$:

$$Q_{12} = \frac{E_{b1}-E_{b2}}{R} = \frac{\sigma\left(T_1^4 - T_2^4\right)A}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1}$$

Result

Per unit area:

$$\frac{Q_{12}}{A} = \frac{\sigma\left(T_1^4 - T_2^4\right)}{\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1} = \varepsilon_{eff}\,\sigma\left(T_1^4 - T_2^4\right)$$

with $\varepsilon_{eff} = \left(\dfrac{1}{\varepsilon_1}+\dfrac{1}{\varepsilon_2}-1\right)^{-1}$. For black plates $\varepsilon_{eff} = 1$ and the result becomes $\sigma(T_1^4-T_2^4)$.