2. Benchmark solution

2.1. Calculation method used for the reference solution

\(T(r,z,\theta )={R}^{2}\mathrm{cos}l\theta\)

with \(l\) Fourier harmonic number

\(-\Delta T=({l}^{2}-4)\mathrm{cos}l\theta =S\)

\(\overrightarrow{\phi }=-(\lambda \overrightarrow{\nabla T})=[\begin{array}{}-\mathrm{2r}\mathrm{cos}l\theta \\ 0.\\ +(\mathrm{lr}\mathrm{sin}l\theta )\end{array}\)

on \([\mathrm{AB}]\) and \([\mathrm{ED}]\): \({\phi }_{0}=\overrightarrow{\phi }\mathrm{.}\overrightarrow{n}=0.\)

on \([\mathrm{BC}]\): \({\phi }_{0}=2R=2.\)

on \([\mathrm{CD}]\): \(\overrightarrow{\phi }\mathrm{.}\overrightarrow{n}=2R=\frac{2}{R}(2{R}^{2}-{R}^{2})=h({T}_{\mathrm{ext}}-T)\)

Hence \(h=\frac{2}{R}=2.\)

\({T}_{\mathrm{ext}}=2{R}^{2}=2.\)

Only the source term varies according to the harmonic \(({S}^{l}(r,z)={l}^{2}-4)\)

In the following models, we will solve the problem on harmonics 1, 2 and 3.

2.2. Benchmark results

Temperatures and flows at points \(B,C,D,F,G\).