Electronics > Projects, Designs, and Technical Stuff
Determining heat dissipation of 3D printed box - final results!
HendriXML:
I think this experiment is delayed due to ordering fake LM335Z's from aliexpress. Off course the dispute period is over.. These weren't even that cheap. What a crap!!
HendriXML:
Being short on temperature sensors I took one LM335Z from another project and used that.
So no controlling a temperature difference, only keeping the heatsink at 60 deg. The room temperature was about 25 deg.
The results give some nice insights.
I measured the voltage across the sense resistors (0.203 ohm combined) with a Fluke in Min/Max mode. The min/max values are in the table, but the avg is the one that matters.
So the best case is no enclosure in which 6.7W can be dissipated. In reality this could be higher if the heatsink temp was allowed to rise.
In the worst case 2.8 W was dissipated. In this case the ventholes where covered by duck tape. Because the top gets quite warm, this will give slightly better results than having no air slots at all.
The box in its intended state can dissipate 3.2W, only slightly better. This proves the idea that the slim slots are obstructive.
However the box with no top at all results also in a significant drop in dissipation capability relative to the no enclosure situation.
The conclusion that I draw from this experiment is that 60 deg vs 25 deg does not produce much air pressure. So it is easily obstructed. To be effective large slots are required.
Maybe I'll redesign the top to test whether that improves things. Or redo part of the experiment with a 80 deg heatsink temperature.
HendriXML:
Will also change Resistor R11 in the PID control to something larger, that will make the (slow) oscillation a bit less. It used to control a large thermal mass, now its a lot less.
Another improvement in the test setup will be to isolate the LM335Z a bit, without isolation the non enclosed situation will probably have a stronger sideways cooling effect on this sensor, so the temp reading might be lower than the actual temp of the heatsink.
The isolation will have a small effect on the cooling capacity of the heatsink.
HendriXML:
I updated the PID control, it is now much more stable (R11 -> 100K)
The sensor got isolation.
I did measure the room temperature with the LM335Z it was 27.8 deg.
The heatsink was regulated to 90 deg. A temperature difference of 62.2 K.
The free air wattage was 11.4 so the heatsink to air had a thermal resistance of 5.46 K/W that's a lot more than the 3.8 K/W (FA-T220-38E) of the datasheet.
https://nl.mouser.com/datasheet/2/303/sink_f_r-1265536.pdf
The difference between covered air slots and uncovered is minimal. I think the top slots are too small. This was one concern when I was designing those slots. Putting the circuit in a topless box, also makes a 27% difference, that is a lot.
With a 90 deg heatsink in the box, the top did become somewhat soft. It has to be said that driving the transistors at 90 deg at this power is most likely not a good idea. I don't know the thermal resistance of the silicone pads (probably rubbish) so I can't be certain about overheating.
HendriXML:
If I look at the specifics of PLA
https://www.makeitfrom.com/material-properties/Polylactic-Acid-PLA-Polylactide
It show that it has a high Specific Heat Capacity (1800 J/kg-K) and a low Thermal Conductivity (0.13 W/m-K). So heating a 100 gr of PLA 62.2K would take 11 kJ.
That is 37 minutes of 5 watt all of the power was only used for heating up, no losses.
What if the most prominent cooling effect of the box came from heat capacity. That would explain the small difference between uncovered vs covered.
So to do this experiment well, I should leave it warming up for at least an hour. Maybe more.
My experiments had some duration, but not in that magnitude.
So I will redo the experiment of the uncovered/uncovered test cases at 80 deg. 90 deg is a bit too hot.
Also covered air-slots might be better than no air-slots, so there usefulness can only be proven not disproven.
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