Introduction & Context

Radiation exchange between parallel gray plates is a canonical problem in thermal process engineering. It quantifies the net radiant heat flux between two large, diffuse, opaque surfaces separated by a non-participating medium. The result is used to size radiant heaters, predict heat losses from furnace walls, design shielding systems, and set boundary conditions for CFD and energy-balance models in reactors, dryers, and kilns.

Methodology & Formulas

Temperature conversion
\[ T_{\text{K}} = T_{\text{°C}} + 273.15 \]
Interchange (effective) emissivity
For two infinite parallel gray surfaces the geometric view factor is unity and the interchange emissivity is \[ \varepsilon_{12} = \frac{1}{\dfrac{1}{\varepsilon_{1}} + \dfrac{1}{\varepsilon_{2}} - 1} \] with the numerical safeguard \[ \varepsilon_{12} = \frac{1}{\max\left(\dfrac{1}{\varepsilon_{1}} + \dfrac{1}{\varepsilon_{2}} - 1,\; 10^{-9}\right)} \] to avoid division by zero.
Net radiation flux
The net radiant heat flux from surface 1 to surface 2 is given by the Stefan–Boltzmann law: \[ q^{\prime\prime} = \varepsilon_{12}\; \sigma\; \left(T_{1}^{4} - T_{2}^{4}\right) \] where \[ \sigma = 5.670 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4} \] Units conversion: \[ q^{\prime\prime}_{\text{kW}} = \frac{q^{\prime\prime}}{1000} \]

Validity Criteria
Parameter	Range	Remark
\(\varepsilon_{1},\ \varepsilon_{2}\)	\(0 < \varepsilon \le 1\)	Emissivity must be strictly positive and ≤ 1.
\(\varepsilon_{12}\)	\(\varepsilon_{12} \le \min(\varepsilon_{1},\varepsilon_{2})\)	Interchange emissivity cannot exceed the smaller surface emissivity.

For infinite or very large parallel plates the net heat flux \( q \) is:

\( q = \sigma (T₁^4 - T₂^4) / (1/\varepsilon₁ + 1/\varepsilon₂ - 1) \)
\( \sigma = 5.670 \times 10^{-8} \) W m⁻² K⁻⁴
\( \varepsilon₁ \) and \( \varepsilon₂ \) are the emissivities of the two surfaces (0 < \( \varepsilon \) ≤ 1)
\( T₁ \) and \( T₂ \) are the absolute temperatures of the plates in kelvin

Insert the values and solve for \( q \); the result is in W m⁻².

A single thin shield (surface 3) reduces the heat flux by roughly one-half if its emissivity is similar to that of the plates:

New thermal resistance becomes \( (1/\varepsilon₁ + 1/\varepsilon₃ - 1) + (1/\varepsilon₃ + 1/\varepsilon₂ - 1) \)
For \( \varepsilon₁ = \varepsilon₂ = \varepsilon₃ \) the resistance doubles, so \( q \) drops to ≈ 50 % of the unshielded value
Multiple shields multiply the resistance, giving \( q \approx q₀ / (N + 1) \) for \( N \) identical shields

No—the simple two-surface formula assumes a non-participating medium. When the gas absorbs/emits (e.g., CO₂, H₂O vapor) you must:

Add a gas radiation term using the gas emissivity \( \varepsilon_g \) and absorptivity \( \alpha_g \)
Couple the surface and gas energy balances iteratively or with view-factor networks
Software such as Hottel charts or CFD radiation models is usually required

Because emissivity appears in the denominator, a small change has a large effect:

Doubling both \( \varepsilon₁ \) and \( \varepsilon₂ \) from 0.2 to 0.4 increases \( q \) by ≈ 60 %
Typical process values: oxidized steel 0.8–0.9, clean aluminum 0.05–0.1, refractory 0.6–0.8
Measure or coat surfaces to control \( \varepsilon \); high-emissivity coatings raise heat removal, low-emissivity shields reduce it

Worked Example: Radiating Roof Slab above a Furnace Hearth

A continuous steel-strip annealing furnace has a flat roof slab (Plate 1) and hearth (Plate 2) that may be treated as infinite parallel gray surfaces. During a holding period the roof is at 250 °C and the hearth at 150 °C. Estimate the radiant heat flux that must be removed by the cooling system to keep the roof temperature constant.

Knowns

Stefan–Boltzmann constant, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
Roof emissivity, ε₁ = 0.80
Hearth emissivity, ε₂ = 0.85
Roof temperature, T₁ = 250 °C (523.15 K)
Hearth temperature, T₂ = 150 °C (423.15 K)

Step-by-Step Calculation

Compute the effective emissivity for two gray, parallel, infinitely large plates: \[ \varepsilon_{12} = \frac{1}{\frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_2} - 1} = \frac{1}{\frac{1}{0.80} + \frac{1}{0.85} - 1} = 0.701 \]
Evaluate the temperature difference term: \[ T_1^4 - T_2^4 = (523.15)^4 - (423.15)^4 = 5.05 \times 10^{10} \; \text{K}^4 \]
Calculate the net radiant heat flux: \[ q'' = \varepsilon_{12} \, \sigma \, (T_1^4 - T_2^4) = 0.701 \times 5.67 \times 10^{-8} \times 5.05 \times 10^{10} = 1703 \; \text{W m}^{-2} \]

Final Answer

The cooling system must remove approximately 1.70 kW m⁻² of radiant energy from the roof slab to maintain steady temperature.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle