Feser's New Lineup

[Kaldskryke]

It's in the equation, but it most likely has less to do with distance. It's probably because the thermal resistance due to the convective boundary, water to copper and brass to air, is much greater than that of the copper to brass. but I'm just speculating... Waterlogged was speculating earlier that the brass to air term that dominates... Maybe feser knows

Don't you mean water to brass and copper to air?

Than homogeneous Al? Why would you average?

...

XS needs a resident thermodynamicist...

I don't mean to take the mean of the two values. I'm suggesting you approximate the system as a (very) simplified one-dimensional geometry that goes: free-stream water -> boundary layer water -> brass tube wall -> solder material -> copper fin -> air boundary layer -> free-stream air. To determine the overall heat transfer coefficient for the system you'll need to work out the convective heat transfer terms separately (Nusselt number differs, etc) but you can join the conductive terms if you weight their conductivities according to their length. It's just a convenient simplification that doesn't change the answer in this kind of geometry, and I was just using it to illustrate that I think that a thin tube wall and a thin layer of solder are less dominant terms than the much longer length of copper fin, even if the solder is particularly resistant.

In reality, the fact that convection occurs along the entire length of the copper fin makes the one-dimensional simplification all but useless for actually doing the math, but I was only intending to invoke it for illustration.

I should maybe dig out my heat transfer textbook from college and see if I can take the simple fin example and change the boundary conditions to fit louvered fins and come up with a more realistic answer, but I don't feel it's worth the effort. Afterall you may be right that the convective terms dominate