Two large parallel plates (emissivity = 0.7) exchange radiation. To cut the net radiant heat transfer by 75%, thin parallel shields (emissivity = 1) are inserted. How many radiation shields are required?

Introduction / Context:Radiation shields reduce net radiative heat exchange between two surfaces by inserting additional parallel surfaces. Each added surface splits the exchange into multiple “resistances in series,” lowering the net transfer. This problem asks for the number of shields needed to achieve a 75% reduction.

Given Data / Assumptions:

• Two infinite, parallel, diffuse surfaces initially exchange radiation directly.
• Shields are thin, parallel, and have emissivity ε = 1 (black).
• Target: reduce heat transfer to 25% of original (i.e., 75% reduction).

Concept / Approach:For black (ε = 1) parallel plates, inserting N black shields divides the space into N + 1 gaps. Because each surface is black, each gap transfers the same heat, and the net becomes Q_new = Q_original / (N + 1). Therefore, to achieve 25% of the original: 1 / (N + 1) = 0.25 → N + 1 = 4 → N = 3 shields.

Step-by-Step Solution:

Verification / Alternative check:This result is a well-known limiting case; for non-black shields, a more general radiation network is used, but the black-surface series model captures the required count here.

Why Other Options Are Wrong:

• One or two shields only yield 1/2 or 1/3 of original, not 1/4.
• Four shields give 1/5, which is more reduction than required.
• “Shields ineffective when ε = 1” is incorrect; with ε = 1, they still divide exchange into series segments.

Common Pitfalls:Confusing emissivity effects with the series segmentation effect; overlooking that black shields still reduce net heat by splitting the exchange.

Final Answer:Three shields