I am using Ironbug to set up a Low Temperature Radiant zone system. The simulation runs and seems to be working correctly but I am getting the following error with regards to the internal source construction:

** Severe ** InitConductionTransferFunctions: Found Material that is too thin and/or too highly conductive, material name = METAL SURFACE
** ~~~ ** High conductivity Material layers are not well supported for internal source constructions, material conductivity = 45.250 [W/m-K]
** ~~~ ** Material thermal diffusivity = 1.157E-005 [m2/s]

Can you share your whole construction and the whole error message?

TL;DR
I think by either changing your material thickness or the time step used in your simulation, you can resolve this message, as described here:

It might not affect your calculation, as you are probably working with very thin layers.

Longer version
There are a few things to clarify that are in your message.

Thermal diffusivity
This becomes your most important factor to look at in transient simulations, similarly as you look at thermal conductivity in steady state.

Thermal diffusivity of materials (\alpha) can be calculated based on the thermal conductivity (\lambda) divided by your material density (\rho) and isobaric specific heat capacity (c_p) of the material.

\alpha = \frac{\lambda}{\rho c_p}

Based on your your 11.57 mm^2/s value, you are using something close to steel in your example.

This doesnât really make sense in a low temperature radiant floor system, so I assume that you are trying to model a wall/ceiling panel. In a floor you would try to flatten its temperature oscillation over time, often by using âheavyâ materials like concrete.

Heavy here doesnât only refer to mass (mass accounting only for the density - \rho), but incorporating the thermal capacity and conductivity as well. Essentially ou are trying to look for materials that have very low thermal diffusivity, as that weird unit [length^2]/s tells you how fast heat travels in a material.

Low temperature metal panels on the other hand (given that enough surface area is exposed to the user) could improve your thermal comfort by controlling the surfacesâ mean radiant temperature, but they could still lose heat quite fast as opposed to e.g. concrete or masonry structures.

From a calculation point of view, few mm thick materials can cause problems (this is why I asked to share the whole message, because it is supposed to give a threshold value to you).

Error message
EnergyPlus is trying to warn you, that you are using too thin layers.
It is using your thermal diffusivity and the timestep to calculate what is your minimum thickness that this time step is able to handle:

IF (Alpha > HighDiffusivityThreshold ) THEN
DeltaTimestep = TimeStepZone * SecInHour
ThicknessThreshold = SQRT(Alpha * DeltaTimestep * 3.d0)

If any of your layers are below this threshold, you are getting thrown this error, that you are getting:

IF (Material(CurrentLayer)%Thickness < ThicknessThreshold) THEN
CALL ShowSevereError('InitConductionTransferFunctions: Found Material that is too thin and/or too ' &
//'highly conductive, material name = ' // TRIM(Material(CurrentLayer)%Name))

Maybe there is a more up to date version of the algorithm, but I am not that familiar with the EnergyPlus source code. The message itself should have a lot more information though, that your question is not including:

Reason behind this error
It has to do with the solver in EnergyPlus that is solving the heat equation: your layer thicknesses define the node spacing (letâs say they are evenly distributed, call them \Delta x) for EnergyPlus constructions to be calculated, and the given node spacing is likely having a small value, for which it cannot produce stable results under a certain time step (\Delta t). (This is a due to having calculation errors with finite methods, a topic which is beyond the scope of this post.)

Or if you reverse it, your defined time steps per hour (I think 6 is the default by EnergyPlus) define your minimum layer thickness. More on the current (?) implementation in this paper (from 2015):

Other references
This post explains the relation between your element size and thermal diffusivity:

The equation written in the answer \left( L= \sqrt{ \frac{kdt}{\rho c_p} } \right) is the same algorithm that you can find code snippet above to define ThicknessThreshold, itâs notation is a bit different (k=\lambda).

Similar post on Unmet Hours, I guess you are trying to do a similar thing: