Prior to hearing this, I would have bet just about any amount of money that aluminized steel had the same thermal conductivity as regular steel.

So would I, and after further thought, I don't think I believe the thermal conductivity number in the AK Steel product data sheet.

Here is why I say that - we can think of the problem in terms of conductive heat transfer:

Fourier's Law of Heat Conduction: q = k A (ΔT/s) [1]

q =heat transfer (W)

k = thermal conductivity of the material (W/m°K)

A = heat transfer area (m

^{2})

ΔT = temperature difference across the material (K)

s = material thickness (m)

This formula can be extended to calculate heat transfer across multiple materials in series:

q = (ΔT) / ((s

_{1}/(k

_{1}A)) + (s

_{2}/(k

_{2}A)) + ... + (s

_{n}/(k

_{n}A))) [2]

We can use these thermal conductivity values:

Low carbon steel = 64 W/m°K

Aluminum = 205 W/m°K

And these thicknesses provided by AK Steel (the extremes that should give the highest k):

Low carbon steel = 0.15in (0.000381m) – the thinnest they offer

Aluminum = 1.2mil (0.00003048m) – the thickest they offer

We’re calculating a constant k for the clad metal, so it doesn’t matter what you use for A or ΔT and long as you use the same values throughout the steps (see [4] below).

Insert the values into [2] and solve for q (remember there are two layers of Al – same thing in [3] when you figure s for the Al-steel-Al clad).

Rearrange [1] to solve for k = (qs)/(AΔT) [3]

Insert q from [1] and the other values into [3] and solve for k = 70.7 W/m°K which is a lot lower than 89 W/m°K. I just don’t see how you can get to 89 with any reasonable set of assumptions. It almost looks like they did a weighted average k which would not be correct.

Inserting [2] into [3] and simplifying yields

k

_{clad} = (s

_{1} + s

_{2} + … + s

_{n}) / ((s

_{1}/k

_{1}) + (s

_{2}/k

_{2}) + … + (s

_{n}/k

_{n})) [4]

which can be used to calculate the thermal conductivity for any n-layered clad material.

Back to the question at hand, if you use a steel thickness of 0.2in, and aluminum thickness of 1.2mil, the increase in thermal conductivity over plain steel is about 0.5 W/m°K which is completely insignificant. The thicker the steel, the less the increase.

I guess this brings back to where we started. Sorry Ron.

CL