Tuesday, 18 March 2008

Hot maths

As the power lost through the stainless steel barrel in my previous post seemed very low I decided to calculate it as a sanity check. I ignored the heat lost from the barrel by convection and radiation. They may be significant now but when I insulate it they shouldn't be.

The outside diameter of the tube is 6.35mm and the bore is 3.6mm, so that gives a cross sectional area of 2.15x10-5m2. Its length is 0.05m. The temperature difference over that length is 240°C-50°C = 190°C. The conductivity of stainless steel is 17 W/mK. So the heat flow is 17 x 190 x 2.15x10-5 / 0.05 = 1.39W. That means it isn't very much compared to the total power required, so that matches my observation.

The amount of heat flowing into the heat sink is therefore 1.39W and it raises its temperature by 30°C, so the heat sink would have to be 22 °C/W. It was just a scrap one I had laying about so I don't have a spec but it is only 70 x 25 x 20mm so that seems in the right ball park.

Sanity checked!


  1. Do you mean 0.046 W/deg C? Your heatsink looks about 5cc, which gives a volumetric thermal efficiency of about 0.01 W/deg C/cc. This is in line with figures I've seen quoted.

  2. Yes well spotted, I meant 22C / W.

    The heatsink volume is 35cc.

  3. I thought about 5cc for the actual metal content.

  4. Maybe, but according to this


    its the total volume that counts.

  5. Hmm, looking at some heatsinks, 22C/W does seem a bit high for one that size. Perhaps a bit more investigation is required.

  6. Mystery solved: the heatsink is not in its optimum orientation. When I turn it on its side the temperature drops to 30C making it about 7.7C/W which is more reasonable. A very dramatic change, the reference I quoted says the difference should only be 70%. One extra effect is that with it in the barrel down orientation, a lot of heat is convected up from the hot barrel below making the heatsink a lot less efficient.