Benchmark solution
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Calculation method used for the reference solution
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:math:`T(r,z,\theta )={R}^{2}\mathrm{cos}l\theta`

with :math:`l` Fourier harmonic number

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

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

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

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

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

 Hence :math:`h=\frac{2}{R}=2.`

 :math:`{T}_{\mathrm{ext}}=2{R}^{2}=2.`

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

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


Benchmark results
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Temperatures and flows at points :math:`B,C,D,F,G`.