Method of images for infinite cylinder between two parallel surfaces, one insulating and the other constant temperature

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?

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