Thermal gradients

Although my last extruder design seems to work, I am not very happy with it. I don't like the little fan because it is noisy, it isn't easy to make and it is not very thermally efficient. The main heat loss is via the stainless steel bolts and from the big flange. The only reason those parts are needed is because the PTFE insulator does not have the necessary mechanical strength and stability.

Some time ago I tried to use stainless steel as the insulator because it is strong, self supporting and withstands high temperatures. That attempt failed because my thermal gradient was too long; the hot to cold transition was about 50mm. The extruder would run for a while but would always jam before an object was completed. Once the stainless steel barrel was fully up to temperature the amount of filament that is soft but not fluid is sufficient to provide too much resistance to be pushed. This theory was confirmed when I tried a soldering iron heater, which also has a long thermal gradient along its length.

I also tried PEEK as the metalab guys had success with it but that seemed to suffer from the same problem.

I had always intended to revisit the stainless steel idea with a shorter transition zone but when I saw Larry Pfeffer's stainless steel extruder it provided an extra idea of thinning down the pipe at the transition. That allows a short transition without too much heat loss or loss of mechanical strength.

I did some experimentation with this test set-up.



I made a heater with an integral aluminium barrel by turning some aluminium bar in my four jaw chuck, the first time I have used it. I used two 12Ω resistors in parallel this time instead of one 6.8Ω. They give a bit more power and possibly lower internal temperature inside the resistors.

I measured the temperature along the tube at 5mm intervals. The thermocouple is slightly smaller than the internal diameter of the tube. The weight of its cable causes the tip to rest on the roof of the tube and the other end rests on the floor of the filament exit. The thermocouple is itself encased in stainless steel, so there will be some heat leaked along it. Hopefully its casing is thin enough to have little effect compared to the much thicker tube it is sampling.

The heater was powered from a bench power supply and the voltage adjusted manually to give around 240°C in the middle of the heater block. That needed 9.4V which is 14W. I can feel substantial heat rising from it so some insulation would make it a lot more efficient. I have got some ceramic fibre kiln insulation for that, another tip I picked up from Larry's blog.


The first test was with a threaded tube with no constriction. The gradient is not far from linear, it falls off faster when hotter due to more convection and radiation from the hot section. If we assume ABS would be soft but very viscous between say 75°C and 125°C we see that it covers 45 to 60 i.e. 15mm.

I then turned a 10mm section of the tube down from 6.4mm to 4.5mm. The internal diameter of the tube is about 3.6mm so that gives a wall thickness of 0.45mm. That made the gradient steeper between 35mm and 40mm but the length of the perceived problem zone gets bigger. I am not sure how Larry gets away without a heatsink. I think he is using thicker pipe so there is a much bigger difference between the conduction of the constricted section and the rest. Also it takes a long time for the problem to become apparent because heat travels slowly down the pipe.

The final test was done with a heatsink attached just above the constriction. The centre of the heatsink only reached about 28°C. The aluminium block I used to connect it got hotter but was still comfortable to touch so less than 50°C.



The gradient between 30mm and 40mm is now much steeper. Odd that it is not between 25mm and 35mm where the constriction is. Almost like there is a 5mm offset in the readings. Anyway the 125°C to 75°C transition is now only about 3mm.

If we assume the temperature difference across the constriction is 210°C - 80°C = 130°C, the conducted heat loss is temperature difference × thermal conductivity × cross sectional area / length. So 130 × 17 × π × ((2.25×10^-3)² - (1.8×10^-3)²) / 10×10^-3 = ~1.3W, about 1/3 of the loss through the bolts and PTFE in my previous design.

So it looks promising, I need to add a nozzle and some insulation and see if it will extrude.