Critical insulation thickness for a cylindrical cable: An aluminium-conductor cable (in air with h = 6 W/m^2K) is insulated with rubber of thermal conductivity k = 0.15 W/m·K. What is the critical thickness (radius) of insulation?

Introduction / Context:
For cylindrical systems (wires, pipes), adding insulation does not always reduce heat loss. Up to a certain “critical radius,” adding insulation can increase external surface area faster than thermal resistance, increasing heat transfer. Correctly computing the critical insulation radius is crucial to avoid counter-intuitive performance.

Given Data / Assumptions:

• External convection coefficient, h = 6 W/m^2K (air).
• Insulation thermal conductivity, k = 0.15 W/m·K (rubber).
• Cylindrical geometry, external convection to air.

Concept / Approach:
For a cylinder with external convection, the critical insulation radius r_c is given by r_c = k / h. If insulation thickness is increased to this radius, heat loss is maximized; beyond it, added insulation reduces heat loss as expected.

Step-by-Step Solution:

Verification / Alternative check:
Dimensional check: k (W/m·K) divided by h (W/m^2·K) → m, consistent with a length scale; typical values for air convection and rubber indeed yield a few centimeters.

Why Other Options Are Wrong:

• 40 mm, 160 mm, 800 mm are not consistent with r_c = k/h for the given numbers.
• 2.5 mm is an order of magnitude too small (would require h ≈ 60 W/m^2K).

Common Pitfalls:
Confusing critical thickness for cylinders with that for spheres (r_c = 2k/h) or thinking any insulation always reduces heat loss; for small wires in low h environments, initial insulation can increase heat loss until r > r_c.

Final Answer:
25 mm