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cdru
I know it's an older post, but your numbers are incorrect for what you're trying to compute as you are confusing thermal conductivity with heat transfer coefficient.
Thermal conductivity is how well heat flows within a material. For our heater blocks, we want heat to flow very well so that it's even and no hot spots right next to the heater but not on the far side of the melt chamber. For this, we choose aluminum or brass because they have a "k value" in the hundreds of W/(m K). But because the thermal conductivity is so good, this is why all metal hot ends heat breaks aren't made of aluminum...it would wick all that heat away up into the cooling fins defeating the purpose of a heat break. So instead they are made of stainless steel which has a k value of around 16 W/(m K). Polycarbonate has a k value of .19 W/(m K). It would make a excellent material for a heat break as it resists the transfer of heat...if only it could withstand the heat (it can't). In terms of your enclosure, 1900 Watts is the rate of heat transfer WITHIN THE POLYCARBONATE if you had the inside surface at 80 degrees Celsius and the outside at 20 degrees Celsius. Slowly over time though the outside surface would raise in temperature as more heat from the inside was generated. That is then where the heat transfer coefficient comes into play...
The heat transfer coefficient is basically how well the material gives off heat, usually in the form of convection between solids and a fluid or gas.
This multiwall polycarbonate sheet like was shown above has a coefficient of 3.7W/(m² K). I found a few solid 6mm sheets that had a coefficient of 3.5W/(m² K). A box with a surface area of 1 meter made of solid polycarbonate 6mm thick would lose 3.5 W/(m² K) * 1m² * 60 Kelvin = 210 Watts, NOT 1900 Watts.
Yeah I guess I didn't explain that well enough.
You are correct in that in a perfectly stagnate room it would be much less than 1900 Watts, but due to convective transfer and the (essentially) infinite mass of the air in the room, this is not the case. The heat transfer coefficient of air is between 10 and 100 W/m^2/K so I'm just assuming that convective air transfer >> conductive transfer through enclosure. My printer is in a large room that people are constantly coming and going from and there is often a fan on or window open or something so this is a
pretty safe assumption. I'm sure if you had it in a closet or small, unoccupied room you would get a much smaller heat transfer coefficient for the air and therefore less energy lost, but you want to plan for the worst case scenario to avoid variability. If I calibrate my enclosure so that 99% of the time the air is stagnate around it and insulation is sufficient, I'm going to get fluctuation that one time that I open the door fast and cool it down a few degrees C. By assuming the "completeness" of the heat transfer to room temp I'm eliminating all external variables. I know my calculations are rough, but I'd bet they're within 50% for a setup like mine.
What you're seeing is the
U value, which is called
thermal transmittance. This is used often in buildings and is used to measure entire walls, or in instances like windows where there is a gas between panes. It is similar to k in that it measures heat flux through the material rather than being a heat transfer coefficient. They give it in different units because it is usually experimentally measured for each system of materials, especially in one like the poly you linked to where the heat flux isn't uniform due to the air pockets. It looks like fantastic stuff and you are very correct in your math for it's heat loss, but for a solid piece of polycarbonate there is no way the U value would get that low. If it has a k value of .19 W/mK, then for a 6m piece the U value would be the k value divided by thickness, or 32 W/m^2K. In that case, (32 W/m^2K)(60K)(1m^2) = 1920 Watts.
But in the same way that my values are exaggerated because of my assumption, so would the 210 watts you calculated which is really cool because that's a really low heat loss!
