# Is there a way to calculate,

Is there a way to calculate, how much thermal cooling a heatsink has to have to keep the heatbreak functional? I designed a couple of hotends and find it cumbersome to build different heatsinks and to try them.

Im wondering if solidworks simulation or similar program might do the trick?

As far as thermal energy throughput goes, it’s a very similar calculation to electrical resistance. You need the heat break’s thermal conductivity, its length and cross-section, as well as the thermal difference you want to achieve.
Then size the heatsink accordingly.

How is the fam annoying @Rene_Jurack ​? My fan is quite and on all the time the printer is on. Is it your particular printer? Or you prefer passive cooling in general?

Too much math after a full day of software engineering at the office​:joy:

@kenneth_rooks I tried that, but maybe I am pushing the wrong buttons…
@Alex_Skoruppa They are fanless How do you know?
@Thomas_Sanladerer Ok. I know the material, I got the drawings of the heatsink, I know delta T is about 250K at max. Currently, my smallest heatsink peaks at 4W cooling till slightly below roomtemperature. At least I can cool a 4W heater to 20 deg, no math involved. An E3D-V6 heatsink does handle 7W but with higher mintemp (roughly 35-40deg) in the same testsetup. I can print with it just fine, but I want to do some math to verify…

@kenneth_rooks I prefer silence in general

Knowing the thermal resistance of a given heatsink (K/W) is probably 95% of any thermal design - figuring out the heatbreak’s performance is trivial compared to that.
You can estimate the thermal resistance of a heatsink on paper, however thermal simulation is probably a lot faster and more accurate.

How are you cooling to below room temperature? That is a magical heatsink!

The heatsink must be hotter than the air for any heat transfer to occur.

As for the original question, it is a trivial calculation if you know the heatbreak materials and dimensions.

@Tim_Elmore1 you’re right, i completely missed that!
@Rene_Jurack no heatsink is able to cool a load below ambient unless it uses evaporative, compressor, peltier etc. cooling.

a.k.a. active cooling

Could be measurement error… I use different thermistors types in the test.

@Thomas_Sanladerer the thermodynamics of heat dissipation is art and science. For example two heat sinks with identical mass and surface area will have vastly different cooling rates. Think of a heatsink like an e3d v6 with horizontal fins, then think of a heat sink of similar mass but thicker fins but shorter. The modulus of thick intersections can have a sizable effect of cooling efficiency.

Or think of the same size heat sink as an E3D, but with vertical fins that allow heat to migrate upwards along the filament path as apposed to the horizontal fins of an E3D. Combine that with how the fins interact with one another when one has a different temperature than the next as you move away from the heat source creating a gradient of Delta-T for cooling. That’s why thermal simulation software is so expensive. It’s a difficult problem to solve.

Don’t forget, the cooling power of a heatsink is roughly linear with how many degrees it rises above ambient. You have to define your allowable temp rise, and then you can evaluate heat sink designs to achieve that. Big vendors like Digikey will have watts per degree kelvin ratings for a particular orientation or orientations of the heatsink.

An all-metal hot end can only have a very low temp rise for PLA to print reliably. For example, I get PLA jamming in one of my printers if the heatsink exceeds around 38C. A PTFE-lined hot end is way more tolerant to high heatsink temps.

Adding to the data: I collected some measurements when evaluating a E3Dv6 water-cooled heatsink. With a normal air-cooled E3Dv6 setup at 190C, the temperature inside the heatbreak at the top of the transition zone was 40.5C, with 24.8C ambient. This is with thermal paste, but no filament movement.

The water-cooled heatsink dropped the heatbreak internal temperature to 31.6C in the same conditions.

Du kannst dir das doch ganz simpel ausrechnen. Du hast die Querschnitts flache deiner Heatbreak und die Temperatur, die dort anliegt. Daraus ergibt sich die Energie, welche dort übertragen wird und abgeführt werden muss.

Nehmen wir als Beispiel mal das Dice spezial. Da habe ich mit 3.7W/s gerechnet, was anhand der Leistung eines 30x30x10mm 12V Lüfters und der vorhandenen angeströmten Fläche des KK abgeführt werden konnte. Dabei wollte ich 28 Grad oberhalb des Einstiches der HB erreichen. Und das ist auch praktisch so. Allerdings nicht, wenn Leute den 5V Lüfter nutzen. Hier werden nur rund 2.9W abgeführt, was nicht ausreichend ist. Man müsste jetzt eben die aktive Fläche des KK vergrößern.

Im Microextruder, als Vergleich, können selbst bei minimalster Durchflussmenge ständig 19W/s abgeführt werden.