Hmm, I think at minimum calculus will need to be involved here. Because we can’t just assume that the heat is spread evenly in the copper - it’ll likely be hotter at the bottom, leading to thermal throttling earlier than expected. On the other hand, there’s going to be heat dissipation into the air, which will help cool the block somewhat
Comment on Copper
KSPAtlas@sopuli.xyz 8 hours agoLet’s assume the dimensions of the copper block are 40mm40mm160mm (I’m not taking the heat spreader into account here)
That results in a volume of 256000mm3, or 256cm3
Copper (at 20C) has a density of 8.935 g/cm3, so that’s roughly 2.28736kg of copper
Copper has a specific heat capacity of 384.603 J/(kg K)
Using E=cm∆t, we can figure out that it would take ≈ 70378J of energy to heat the copper block to 100C, starting at 20C
With a TDP od 120W, that means it would take 586 seconds to heat the block to 100C, or 9m46s
This is probably way off but I was bored
Contramuffin@lemmy.world 7 hours ago
Einskjaldi@lemmy.world 5 hours ago
The conduction in copper is fast enough that there’s not much of a difference between the top and bottom.
Contramuffin@lemmy.world 4 hours ago
Copper conductivity is fast, sure, but it’s not fast enough to have equal temperatures at the top and bottom for such a big chunk of copper. That does affect the time to thermal throttle pretty significantly, actually. If we assume completely homogeneous temperatures across the block (ie, instantaneous heat transfer), according to my model, it’ll take 703 seconds to thermal throttle. With heat transfer, the time drops to 592 seconds - a difference of almost 2 minutes
Eheran@lemmy.world 4 hours ago
Heat transfer will not limit much, but heat loss should add a significant amount of time. How did you model that?
Contramuffin@lemmy.world 4 hours ago
I left another comment going into more detail about the model specifications, if you’d like to read into it. But briefly: I took the copper heat conductivity coefficient and the air heat transfer coefficient. I sliced the copper block into thin slices and modeled heat transfer between each slice, as well as heat transfer between each slice and the surrounding air.
It seems that both heat transfer and heat loss do actually matter quite significantly, but they just cancel each other out almost entirely.
If we assume instantaneous heat transfer, thermal throttling time goes up from 592 seconds to 703 seconds (almost 2 minute difference).
If we assume no heat loss to the air, thermal throttling time goes down from 592 seconds to 500 seconds (about 1.5 minute difference).
Eheran@lemmy.world 2 hours ago
If they cancel out, the system would be in balance and not get hotter. So some thing does not add up. What heat transfer coefficient did you use and which other numbers etc.?
falcunculus@jlai.lu 7 hours ago
Alright, what about a 4040160mm block of copper at 100C ?
potpotato@lemmy.world 8 hours ago
Account for convective loses into air?
m0darn@lemmy.ca 5 hours ago
I got ~32 watts of convective heat transfer using assuming a uniform surface temperature of 99°C and a case fan speed of 0.2 m/s
YerbaYerba@lemmy.zip 7 hours ago
Its going to radiate a little bit too.
SleeplessCityLights@programming.dev 8 hours ago
Your napkin math is the best we have. We will make all decisions based on it.
Napster153@lemmy.world 3 hours ago
They will have entire hard drives explaining KSP Atlas’s shitty math in 3 thousand years…