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As most people who've had any elementary heat transfer course are aware, when insulating a pipe, wire, etc, there is a critical thickness for the insulation below which it causes greater heat transfer than no insulation would. In actuality, since the critical radius is a minimum, one needs to go even farther before it catches up to the uninsulated case. This effect is due to the thermal resistance from the added insulation (a function of material and thickness) being less than the decrease in thermal resistance to convection brought about by the increased surface area due to the thicker insulation. A more detailed explanation is found here

What I wonder is how large this critical thickness typically is for realistic cases such as an electrical wire, a steam pipe, or a water pipe (all cases in STP air)? I realize exact answers aren't possible given the fact that it will vary with material type, temperature diff., and size.

The motivation for this is that it theoretically could prevent wasting insulation or provide a practical avenue for energy dissipation, but Fundamentals of Heat and Mass Transfer 6e by Incropera et al. indicates that the effect is usually too small to be of practical importance.

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    $\begingroup$ You link offers everything you need to compute the answer from a set of known heat transfer properties and computes a estimated largest critical radius of about 1 cm. The tricky parting is finding the correct value of $h$ for a given situation, and engineers write whole books on that problem. $\endgroup$ Oct 13, 2011 at 14:17
  • $\begingroup$ @dmckee I'm fully aware how to do this and I have tables of acceptable values. Ultimately, I'm lazy/busy and hoped that it's already been done so I wouldn't have to reinvent the wheel. If I needed this information on a tight schedule, I'd do exactly what you propose using extreme h values. But I have no pressing reason to do this soon. $\endgroup$ Oct 13, 2011 at 21:43
  • $\begingroup$ In cases of electric heating cables, the thermal insulation is below the critical value, to enable heat "loss" while keeping the cable temperature at a reasonable and safe level. $\endgroup$
    – user15517
    Oct 30, 2012 at 6:51

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Using this equation also found here (under 'Insulation of cylinders'): $$r_{cr, cylinder} = {k\over h}$$ this reference calculates a maximum for critical thickness, when $k =0.05W/m^2K$, and $h = 5W/m^2K$, of $10mm$, which seems widely applicable. Typical k and h values mean that the real value will usually be much less than $10mm$. The effects of radiation and forced convection both decrease the critical thickness value even further.

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