**Newtons law of cooling is likely too over-simplified to give an accurate result, especially over a wide variation in temperatures. It assumes that each unit area transfers the same amount of heat (And also that h isn't affected by T).**

The thermal exchange coefficient h is affected by T, but in an indirect way.

First of all the Rayleigh number is calculated with :

The volumetric thermal expansion coefficient

The kinemathic viscosity

The thermal diffusity

All three are dependent on the temperature.

Then using the Raleigh number to calculate h, we also multiply by k, the thermal conductivity of air, also temperature dependent. Thus the calculated value of h is largely depent on temperature.

But I agree that we work with an average temperature and that h over the area is constant.

That is indeed a simplified approach, which however gives for situations which have been researched many times in literature, such as horizental tubes, adequate results.

An other approach would be to use CFD (Computional Fluid Dynamics) software to calculate the convective heat transfer of a part of the rod for a given temperature. Additional we can then also calculate h.

I have CFD software, but have still to learn how to use it beyond the basics. But I have seen in literature several quite complex simulations with CFD software which give errors of 1% to 2 % compared to the real measured data. This because CFD is not using approximations in calculating the convective heat transfer coefficient, but is instead solving the differential equations governing the heat transfer. So maybe this is the way to go.

**Instead of using h = Q/(A(Ts-Ta)), the more accurate method is to calculate an average h (taking the shape of the object into account) - done by first working out the Prandtl, Raleigh & Nusselt numbers for the system.**

Using h = Q/(A(Ts-Ta)) is using the forumula as used by the Lugano team in a reverse way using their data. As such that should give the exact value of h they used.

**Can't paste the formula's for those easily, but this should give you what's needed:**

I had the link myself, but must concede that I did not look at the contents lately.

The formula showed in the link is somewhat different from the one the Lugano team and I used.

If time allows it migh be an idea to use the formula in the link and see how much the data differs.

But there are many formula's for calculating the convective heat transfer coefficient of a horizental tube (or round heated wires).

The formula used by the Lugano testers is normally used for much larger Rayleigh numbers then those valid for the Lugano case.

As such it might be worth the effort to find a better (more accurate) formula for the lower Rayleigh numbers.