I'm trying to conduct a thermal conductivity test. I've set up an experiment where I take a rod of a given material (steel, for example), measure its cross section, length, extreme temperatures and heat conducted and thereby obtain its thermal conductivity.

My problem is that I've tried using a 33Ohm, 5W ceramic resistor to send heat through the test element, but the resistor will heat to about 60 or 70 °C, and the hot side sensor (and the test element) won't heat up to anywhere near that temperature.

Has anybody dealt with the problem of getting the heat from the resistor to be conducted in a specific direction, instead of dissipating on the air?

_I've been using a thermal grease film of around 1W/K to create contact between the test element and the resistor.

_I haven't applied presure between the resistor and the test element.

_I haven't used a vacuum pump yet (that's the plan going forward)

  • \$\begingroup\$ Thermal grease isn't going to cut it if you don't apply pressure as well. \$\endgroup\$ – Hearth Jun 10 '18 at 19:56

I don't think that this is the place for such question which is rather a physics question but since is here I will try to answer using the few info given.

Looking at the size of a 5 W ceramic resistor I can just guess that either it is a big rod, either you have a small contact area with it

The iron has a thermal conductivity 10 times higher than ceramic

The iron has a specific heat capacity of 444J/kgC , that means that with 1W, even all energy goes to the iron rod, you need 444 seconds to heat on kg with one Celsius degree, for 5W you need 88s, for a 250g rod 20s. Only with one degree with perfect thermal contact and no cooling at all.

Also the contact area is to small and all the heat is absorbed by the rod and dissipated into the air to fast compared with the exposed area of the resistor.

To show my point see the following schematic:


simulate this circuit – Schematic created using CircuitLab


TRc1 and TRc2 are thermal resistance of the ceramic from the core to the rod surface / from the core to the exposed surface of the resistor where TRc1 > TRc1 because the contact area with the rod is much smaller

TRair 1 and TRair2 are the thermal resistance from the rod surface to the air / from the resistor surface to the air where TRair2 > TRair1 because the area of the rod is much bigger than the exposed area of the resistor.

TRrod is the thermal resistance of the rod, for iron ten times lower than ceramic.

So you will have a much better thermal conductivity from the ambient temperature to the rod contact point while also a better thermal conductivity from the heater core to the finger.

Now , what to do ?

Use a better suited heater , some that has a metal surface for attaching a heat sink where you can attach the rod like this one or use a MOSFET suited for a heat sink.

Put a thermal insulation on the the area not in contact with the rod to minimize the loses if you need energy calculations.

Make the rod with a large flat end to attach the heater for better thermal contact.

It would be useful to have a rough calculation of the power needed to heat the rod , my opinion is that 5W is not enough.


You do not include enough pertinent information. The surface area between the iron and resistor and photo of your test setup would help. Specific type of ceramic and iron.

Try using a C-Clamp to apply pressure.
In this image I used C-Clamps to hold a CoB LED mounted on a small plate of copper to a copper water pipe.

enter image description here

I used CoB LEDs mounted to a rectangular bar.
This image is two 30 watt CoBs mounted on a copper bar.
I was testing conductive heat transfer and the influence of free convection.

enter image description here


To test the thermal properties of a metal rod I would be looking at two parameters.

Thermal resistance

  • the length / cross-sectional area ratio increases resistance, so measure this.
  • external ambient cooling causes measurement error, so the rod must be well insulated with no heat loss between 2 temp sensing displacements
  • like a Kelvin Resistor test method , the two Test points are away from the two ends where one end is heated
  • I would use magnet wire wrapped around one end then sense Temp after insulation otherwise heat loss will result in errors. The far end must also be sensed away from end unless equally insulated.
  • you must define your accuracy spec to determine all sources of error and thus temp rise to achieve this

Heat Velocity. mm/s per unit length

  • This also depends on L/A dimension ratio (just like Inductance)
  • like electrical terms, it can defined from 0 to 63% rise in final temp or 10% to 90%

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