Critical thickness (radius) of insulation — basic relation. Evaluate the statement: “The critical thickness of insulation equals the ratio k / h0, where k is the insulation thermal conductivity and h0 is the external convective heat-transfer coefficient.”

Introduction / Context:
The concept of critical insulation radius explains why small-diameter pipes can initially lose more heat when thin insulation is added. The classic relation connects the insulation conductivity to the surrounding convection coefficient.

Given Data / Assumptions:

• Cylindrical geometry with external convection.
• Steady, radial conduction through the insulation.
• Constant k and uniform external h0; radiation neglected.

Concept / Approach:
For a cylinder, differentiating the heat-loss expression with respect to the outer radius shows a maximum at the so-called critical radius r_c. The derivation yields r_c = k / h0. For spheres, the corresponding result is r_c = 2k / h0. The statement given matches the cylindrical case, which is the standard scenario in piping.

Step-by-Step Solution:
Total thermal resistance R_total = ln(r_o/r_i)/(2 * pi * k * l) + 1/(h0 * 2 * pi * r_o * l).Heat loss Q = (T_i − T_∞) / R_total.Set dQ/dr_o = 0 → r_c = k / h0 (for cylinders).Thus, at r_o = r_c the heat loss is maximum; beyond this, added insulation reduces heat loss.

Verification / Alternative check:
Textbook examples for pipe insulation consistently report r_c = k / h0 for cylinders and r_c = 2k / h0 for spheres, verifying the formula.

Why Other Options Are Wrong:
(b) contradicts the established cylindrical relation. (c) is incorrect: spheres use 2k/h0, not k/h0. (d) radiation is not part of the classical derivation. (e) Biot number pertains to solids’ internal conduction versus surface convection, not the insulation critical radius.

Common Pitfalls:
Applying the cylinder formula to spheres or plane walls; geometry matters. Also, forgetting that the “critical thickness” here refers to radius, not linear slab thickness.

Final Answer:
True