I am trying to attach the shaft of a brass heating tip to a PLA component. My problem is that the tip will have to reach a temperature of about 200°C and the PLA can only handle a temperature of about 60°C before losing structural strength.

My plan is to first to first cover the shaft with PTFE tubing and then screw the PLA component on top of it. The PTFE tubing will thus have to provide enough insulation to keep the PLA cool. I am now trying to calculate how thick a layer of PTFE tubing will be required to achieve this.

I have thus far figured out that PTFE has W(m.K) thermal conductivity rating of 0.25. I am assuming that the PLA is at a room temperature of 25°C resulting in a temperature difference of 175°C. This apparently means that if I have a $5\,mm$ thick piece of PTFE that I will have $8750\frac{W}{m^2}$ of heat transfer. This about as far as I can come with the maths. I could not find a refernce to what I should after this to calculate how hot the PLA will become.

I am guessing you have to take the heat capcity to calculate the temperature gain per minute, but that doesn't make sense to me because that would mean that it would continue heating until it melts no matter how insulated it is, could that be? I am guessing it should stop getting a hotter once the rate at which it recieves heat and rate rate at which it can dissipate that heat to the air match, but I have now idea what that rate is.

Could any of you possibly help me to get this answer and explain to me how I should calculate it myself next time?


I have now calculated that if I have 110 squre mm of contact between the ptfe and brass (10mm length), and if I have 2.5mm thick PTFE then I would require 5m thick PLA with a length of 70.52 mm (4100 square mm of contact with air) to keep the PLA temperature at 60C.

I am not sure if I calulated the heat exchange between the PLA and air correctly. I first determined that there would pass 1.54w of heat from the brass through the PTFE to the PLA. I calulated it like this:

0.25*140/0.0025*0.00011 = 1.54w

The part I am worried about though is where I calulated how long piece of PLA would be needed. I assumed to calulate this you simply take the heat exchange rate of PLA and doing the same math equation as above but this time make the wattage a constant and the surface area a variable. I did it like this

1.54 = 0.035*35/0.005*x
1.54/x = 24
x = 0.0041 m2 = 4100mm2

I then calulated the circumference of 5mm thick pla and determined that I then need a length of 70.52mm of PLA. Is it correct to asume that heat would spread and disspitate ove the entire surace area like this?

My big question now is did I calulate the amount of heat transfered to the air correctly or is there some other rule of thumb I can use to get a basic idea?

Here is a layer illustartion:


The centre part is the hollow inside of the brass tubing leading to it's end.

  • $\begingroup$ Is there any chance you could post a quick sketch of a diagram? It's hard to work out the geometry from what you've written. (Just post a link to an image and someone else can edit it into your question.) $\endgroup$ – Nathaniel May 5 '13 at 15:11
  • 1
    $\begingroup$ You're probably right that "it should stop getting a hotter once the rate at which it recieves heat and rate rate at which it can dissipate that heat to the air match", but the problem is that calculating convective dissipation tends to require specialised modelling software, so you'll probably have to resort to measurement. $\endgroup$ – Nathaniel May 5 '13 at 15:12

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