Radiation exchange between two infinite parallel plates For two infinite parallel planes with emissivities ε1 and ε2, what is the correct interchange factor (radiation network factor) multiplying σA(T1^4 − T2^4)?

Introduction / Context:
Radiative heat transfer between large, parallel, diffuse–gray plates is a textbook case for using the radiation network method. The net exchange depends on a combined “interchange factor” that accounts for both emissivities and view factors (unity here).

Given Data / Assumptions:

• Two very large (infinite) parallel plates facing each other.
• Diffuse–gray surfaces with emissivities ε1 and ε2.
• View factors F12 = F21 = 1; surroundings do not participate.

Concept / Approach:
For two infinite parallel plates, the net radiative heat flow is Q = σ A (T1^4 − T2^4) / R_rad, where R_rad is the radiation resistance. The equivalent interchange factor, often denoted as 1/R_rad, is 1 / ( (1/ε1) + (1/ε2) − 1 ). This comes from series addition of surface resistances 1/ε − 1 and accounts for multiple reflections between the plates.

Step-by-Step Solution:
Write network: R_total = (1 − ε1)/(ε1 A) + 1/(A) + (1 − ε2)/(ε2 A).For infinite parallel plates, geometric resistance (1/A) simplifies to 1/A since F12 = 1, leading to the classic denominator form.Invert R_total/A to get the interchange factor: 1 / ( (1/ε1) + (1/ε2) − 1 ).

Verification / Alternative check:
Check limiting cases: if ε1 = ε2 = 1 (black bodies), factor becomes 1, so Q = σ A (T1^4 − T2^4), as expected.

Why Other Options Are Wrong:
(b) is an algebraic convenience used in some reflectivity relations but not the correct network factor. (c) and (d) ignore multiple reflections. (e) lacks the proper reciprocal structure and does not reduce to unity for black bodies.

Common Pitfalls:
Forgetting that surface resistances add in series, which introduces reciprocals and the “−1” term.

Final Answer:
1 / ( (1/ε1) + (1/ε2) − 1 )