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?

  • 1
    $\begingroup$ The BC for an insulating plane is $dT/dz = 0$. $\endgroup$ – Jeffrey J Weimer Jan 10 '20 at 14:59
  • $\begingroup$ so i would place like-signed image charges on either side of insulating plane? $\endgroup$ – phdmba7of12 Jan 10 '20 at 15:03
  • 1
    $\begingroup$ 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? $\endgroup$ – Jeffrey J Weimer Jan 10 '20 at 15:07
  • $\begingroup$ mirroring opposite signs will generate constant temperature surface $\endgroup$ – phdmba7of12 Jan 10 '20 at 15:10
  • $\begingroup$ So what happens when you mirror equal signs? $\endgroup$ – Jeffrey J Weimer Jan 10 '20 at 16:47

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.