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Trying to solve for temperature distribution of a infinite cylinder of radius $R$ of uniform time independent heat generation [$P$] = W$\cdot$m$^{-3}$ suspended between two planes a distance $d$ apart, one being at constant temperature $T$ and the other an insulating boundary. The medium of the cylinder and space between planes has thermal conductivity [$\kappa$] = W$\cdot$m$^{-1}$K$^{-1}$ with differential equation being $\nabla^2T=-P/k$.

Was able to use method of images for case of single constant temperature plane, by merely placing second cylinder of negative heat generation density $-P$ an equal distance below.

How would I use image method for the constant temperature/insulating surfaces cylinder sandwich?

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    $\begingroup$ The BC for an insulating plane is $dT/dz = 0$. $\endgroup$ Jan 10, 2020 at 14:59
  • $\begingroup$ so i would place like-signed image charges on either side of insulating plane? $\endgroup$ Jan 10, 2020 at 15:03
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    $\begingroup$ Are you certain that your first case of mirroring the power source on the two sides of the plane is NOT the $dT/dz$ solution, since the profiles generated are symmetrical about the plane? $\endgroup$ Jan 10, 2020 at 15:07
  • $\begingroup$ mirroring opposite signs will generate constant temperature surface $\endgroup$ Jan 10, 2020 at 15:10
  • $\begingroup$ So what happens when you mirror equal signs? $\endgroup$ Jan 10, 2020 at 16:47

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The mirror plane criteria to set a constant temperature is $T = \mathrm{constant}$. Setting a source on one side and an equal sink on the other can generate this.

The mirror plane criteria for an insulated surface is $dT/dz = 0$. Setting two equal sources on opposite sides can generate this.

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