OK, so I have found some time to crunch through this in more detail.
The furnace cavity is cylindrical with diameter, D = 0.1 m, and length > D (more than 10 times D). The furnace is held at constant temperature, T_{env}= 873 K. A spherical sample of radius = 0.0125 m at temperature, T_{samp}(t=0)= 293 K is "instantaneous" introduced into the furnace with no heat loss.
So the Rayleigh Number, \mathrm{Ra}, (when \mathrm{Pr}=1) is given by:
\mathrm{Ra} = \mathrm{Gr}= \frac{g \beta} {\nu^2} (T_{env} - T_{samp}) D^3
To first order I assume that \beta and \nu are temperature independent and values for air at 873 K are used (1.17e-3 K^{-1} and 98.65e-6 m^2s^{-1}, respectively)
Following the equation for \mathrm{Nu}, the heat transfer coefficient, h can be determined by:
h=\frac{\mathrm{Nu}k}{L},
where k is the thermal conductivity of the sample (taken to be ~2.5 W.m^{-1}.K^{-1} for a basalt sample). Should this be for air?
L is the characteristic length scale, here I use the ratio of the sample volume/area \left(L\approx{\frac{V}{A}}\right).
The solution to the lumped capacity method is given by:
T_{samp}(t) = T_{env} + (T_{samp}(0) - T_{env}) \ e^{-r t},
where r=\frac{hA}{C_p\rho{}V}.
My sample is a basalt so...
The mass specific heat capacitance, C_p=1200 J. K^{-1}. kg^{-1}
and the density, \rho{}=2800 kg. m^{-1}
Again, these are assumed to be invariant of temperature.
Still following me? I'll admit it's interesting learning this stuff!
Using the above I calculate Ra(t=0) = 685730 and h (t=0) = 35.9.
By t = 1000 s, Ra(t=1000) = 114221, h (t=1000) = 24.8, and T_{samp}(t=1000) = 776.5 K.
And by t = 2000 s Ra(t=2000) = 38227, h (t=2000) = 20.0, and T_{samp}(t=2000) = 840.8 K.
Since Ra, and hence r, depend on T_{samp} I evaluate T_{samp}(t_n) using r(t_{n-1}) and use 50 s time steps. This will introduce an error, which will be reduced as the time steps decrease, correct?
Are these reasonable values? They seem OK to me. But am I using the appropriate values? By that I mean, am I making the correct distinction between fluid (air) and sample (rock) properties in the equations?
I know I skipped ahead of the simple example you suggested, but I find it easier to learn by applying methods to more realistic situations, assuming of course I have done it correctly.
If I am on the correct track then I will build in the temperature variation of the "constants" (\beta and \nu). Should these be evaluated at
T_f=\frac{T_{samp}+T_{env}}{2}?
Also, in the strictest sense, the furnace cavity is cylindrical (so is the sample) and \mathrm{Pr}\neq1. Would it then be more appropriate to calculate h using:
h = \frac{k} {D}\left({0.6 + \frac{0.387 \mathrm{Ra}^{1/6}}{\left[1 + (0.559/\mathrm{Pr})^{9/16} \, \right]^{8/27} \,}}\right)^2
Cheers for all the help so far!