Re: Enclosure math- why they're so important and how to make one.
April 07, 2015 11:59PM
Quote
Djkingsley
60 degrees Kelvin is -213.15 degrees Celsius. I believe you meant to use 333.15 degrees Kelvin which would give you 3.5 W/m^2K * 1m * 333.15 K = 1166 W
No. The formula is for the Δt...I just happened to put what the Δt was and not (80 degrees C - 20 degrees C). The original example was 20 degree C ambient air temperature and a theoretical 80 degree C build chamber. That's 60 degrees C. If it was in K, it would 293.15 K ambient and 353.15 K build chamber. And look, it's still a difference of 60 just in Kelvin. Kevin...degrees C...when talking about a Δt it doesn't matter between the two.
Re: Enclosure math- why they're so important and how to make one.
April 08, 2015 11:12AM
I have not completely enclosed my delta just yet. However just draping some clothing over most of the openings I managed a temp of 40C with a 100C bed and a 150w heater. It took about 15 minutes to reach 40c with gaping holes. I insulated my printer with a thin layer of cork material. This was at 20c ambient temp. If the temps become too erratic I will add another layer of cork insulation to the sides.

Here is the build blog on the subject. -> [engineerd3d.ddns.net]

Once I get a proper door and insulate the remaining parts, I am confident I can reach the desired 60c with relative ease. Should help with the bed heat up times as well. One more thing, having a blower fan ontop with a charcoal filter has reduced the ABS fumes quite a bit once the print is done, I use the blower to evacuate the printer chamber and filter the air.

-Bruno


My Personal Blog. Build blog.
[engineerd3d.ddns.net]

Modicum V1 sold on e-bay user jaguarking11
Re: Enclosure math- why they're so important and how to make one.
April 08, 2015 11:41AM
This is the quick-and-dirty arrangement I knocked up to improve ABS print quality on the Mini Kossel:



I already had the carboard draught shield in place, so I just put 2 plastic bags over the top of the machine and taped the inner one to the cardboard. The air space between the plastic bags stops the heat escaping too quickly.

Edited 1 time(s). Last edit at 04/09/2015 03:12AM by dc42.



Large delta printer [miscsolutions.wordpress.com], E3D tool changer, Robotdigg SCARA printer, Crane Quad and Ormerod

Disclosure: I design Duet electronics and work on RepRapFirmware, [duet3d.com].

Re: Enclosure math- why they're so important and how to make one.
April 08, 2015 10:31PM
Thanks for the tips and tricks on this, Time to make a test version with: Cardboard box with 2.5mm polystirene sheet fixed to 4 sides and a plexiglass at the front. thanks for the tip regarding silver aluminium IR sheeting like milar with bubble wrap in it, would be awesome in conjunction with IR lightbulbs.
Re: Enclosure math- why they're so important and how to make one.
April 09, 2015 02:03AM
Since this thread has some recent action I thought I would give an update to my build (first page). I have been using my enclosure without any additional heater for months now and I have had no problems with ABS. I have not measured the actual temperature inside when warmed up but it's quite a bit hotter than the room temperature and the only thing heating it up is the hotend (set at 250c) and the heated bed 300X300 set to 85c. I don't think it's getting up to 60c but I have had no problems at all with ABS (using PEI bed so that takes care of adhesion) but if I try to print with the enclosure open (whole front off) then ABS always warps and cracks.
Re: Enclosure math- why they're so important and how to make one.
April 09, 2015 12:10PM
Quote
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!
Sorry, only registered users may post in this forum.

Click here to login