wrote:

... if the metal is a good conductor, eg steel with 50 Btu/h-ft-F, with
poor airfilm conductances on both sides, eg 5 Btu/h-F-ft^2, thinner steel
won't help much. How much, in this case, starting with 0.050" steel?

Heh, don't try to obfuscate the facts by spewing a bunch of calcs as
usual, trying to cover up...

It's 300-year-old physics :-) What's the answer to this simple problem?

Still no clue? Rewrite the steel conductivity as 50 Btu-ft/h-ft^2-F...

Still stuck? Try 600 Btu-inch/h-ft^2-F.

Another clue: the thick steel conductance is 600/0.050 = 12,000 Btu/h-F-ft^2.

And if we halve the thickness, it becomes 24,000 Btu/h-F-ft^2. Wow!

So the steel thermal resistances are 1/12K and 1/24K h-F-ft^2/Btu.

Now what do we do with resistors in series?

Nick