Critical insulation thickness for spheres For a spherical insulation on a hot or cold body, the critical radius (or thickness) is:
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Ak / h0
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B2k / h0
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Ch0 / k
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Dh0 / (2k)
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Ek * h0
Answer
Correct Answer: 2k / h0
Explanation
Introduction / Context:Adding insulation does not always reduce heat loss immediately for cylindrical and spherical geometries. The concept of a critical radius explains why heat loss can first increase before decreasing as insulation thickness increases.
Given Data / Assumptions:
- External convection with heat-transfer coefficient h0 at the insulation outer surface.
- Insulation thermal conductivity k is uniform.
- Steady, radial heat transfer; contact resistances neglected.
Concept / Approach:For spheres, the critical outer radius r_crit is 2k/h0. If the actual outer radius is below r_crit, adding insulation can increase heat loss because the increase in outer area (boosting convection) outweighs the added conduction resistance. Beyond r_crit, further insulation reduces heat loss.
Step-by-Step Solution:Write total thermal resistance R_total(r_o) = R_cond,insulation + R_conv,outer.Differentiate heat loss Q with respect to r_o and set dQ/dr_o = 0 for extremum.Solve condition to obtain r_crit = 2k / h0 for a sphere (compare with k / h0 for a long cylinder).
Verification / Alternative check:Dimensional check: k (W/m·K) divided by h0 (W/m^2·K) gives metres; multiplying by 2 gives a length, as required.
Why Other Options Are Wrong:(A) is the cylindrical result, not spherical; (C) and (D) invert units; (E) has incorrect dimensions.
Common Pitfalls:Forgetting geometry matters: cylinder → k/h0; sphere → 2k/h0. Also, “critical thickness” is often described via the outer radius, not the material thickness alone.
Final Answer:2k / h0